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Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability
2nd PUC Maths Probability One Marks Questions and Answers
Question 1.
Given that E and F are events such that P (E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find p(E/F) and p(E/F).
Answer:
It is given that P(E) = 0.6 P(F) = 0.3 P(E ∩ F) = 0.2
Question 2.
Compute P(A/B), if P(B) = 0.5 and P(A ∩ B) = 0.32.
Answer:
It is given that P (B) = 0.5 and P(A ∩ B) = 0.32
Question 3.
If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4, find
(i) P(A ∩ B)
(ii) P (A/B)
(iii) P(A ∪ B).
Answer:
(i) Given, P(A) = 0.8, P(B) = 0.5, P(A/B) = 0.4
(iii) From the relation P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.8 + 0.5 – 0.32 – 0.32 = 0.98.
Question 4.
Evaluate P(A ∪ B), if 2P(A) = P(B)\(\frac { 5 }{ 13 }\) and p(A/B) = \(\frac { 2 }{ 5 }\).
Answer:
Question 5.
If P(A) = \(\frac { 6 }{ 11 }\), P(B) = \(\frac { 5 }{ 11 }\) and P(A ∪ B) = \(\frac { 7 }{ 11 }\)
(i) P(A ∪ B)
(ii) P(A/B)
(iii) P(B/A)
Answer:
Question 6.
If P(A) = \(\frac { 3 }{ 5 }\) and P(B) \(\frac { 1 }{ 5 }\), find P(A ∩ B), if A and B are independent
events.
Answer:
It is given that P(A) = \(\frac { 3 }{ 5 }\) and P(B) = \(\frac { 1 }{ 5 }\)
As, A and B are independent events, therefore
Question 7.
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the
Answer:
There are 26 black cards in a pack of 52 cards.
Required probability = P (E and F) = P(E) P(F)
Where. E first card drawn is black, F : second card drawn is black
2nd PUC Maths Probability Two Marks Questions and Answers
Question 1.
A coin is tossed three times
(i) E : head on third toss F : head on first two tosses
(ii) E : atleast two heads F : atmost two heads
(iii) E : atmost two tails F : atleast one tail
Answer:
When a coin is tossed three times, the sample spaces S
contains 23 = 8 equally likely sample points.
∴ S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Here, E : set of events in which head occurs on third toss and
F : set of events in which heads occur on first two tosses,
E = {HHH, THH, HTH, TTH} and F = {HHH, HHT}
⇒ E ∩ F = {HHH}
(ii) Here, E : set of events in which atleast two heads occurs and
F : set of events in which atmost two heads occur.
∴ E = {HHH, HHT, HTH, THH}
[Here, we consider the cases of two and three heads, as the condition given is atleast not atmost]
F = {TTT, THT, TTH, HTT, HHT, HTH, THH}
⇒ (E ∩ F) = {HHH, HTH, THH}
(iii) Here, E : set of events in which atmost two tails toss and
F : set of events in which atleast one tail occurs.
∴ E = {HHH, HHT, HTH, THH, HTT, THT, TTH}
[Here, we consider the cases of the two tails, one tail and no tail because the condition is atmost not atleast]
and F = {TTT, THT, TTH, HTT, HHT, HTH, THH}
[Here, we can consider the cases of one or more tails because the condition given is atleast not atmost.]
⇒ (E ∩ F) = {HHT, HTH, THH, HTT, THT, TTH}
Question 2.
Two coins are tossed once, where
(i) E : tail appears on one coin F : one coin shows head
(ii) E : on tail appears F : no head appears
Answer:
If two coins are tossed once, then the sample space S = {HH, HT, TH, TT}
(i) E = Set of events having tail on one coin = {HT, TH},
F = Set of events having one coin showing head = {HT, TH} and
(ii) E = Set of events having no tail = {HH},
F = Set of events having no head = {TT} and E ∩ F = Φ
Question 3.
A die is thrown three times,
E : 4 appears on the third toss
F : 6 and 5 appears, respectively on first two tosses.
Answer:
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216.
The sample space is S = {(x, y, z): x, y, z ∈ {1,2,3,4,5,6}}.
Here, E : set of events in which 4 appears on the third toss and
F : set of events in which 6 and 5 appears respectively on first two tosses.
Question 4.
Mother, father and son line up at random for a family picture
E : son on one end F : father in middle
Answer:
If mother (m), father (f) and son (s) line up for the family picture, then the sample space will be S = {mfs, msf, fms, smf, sfm} ⇒ n (S) = 6 which contains 6 equally likely sample events.
Here, E = set of events having son on one end and
F = set of events having father in the middle.
E = {mfs, fms, smf, sfm}, F = {mfs, sfm} and EnF = {mfs, sfm}
⇒ n(E) = 4, n(F) = 2, n(E ∩ F) = 2
Question 5.
A black and a red die are rolled.
(i) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(ii) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Answer:
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S = 6 × 6 = 36 (equally likely sample events)
(i) Let E : set of events in which sum greater than 9 and
F : set of events in which black die resulted in a 5
E = {(6, 4), (4, 6), (5, 5), (5, 6), (6, 5), (6, 6)} n(E) = 6
and F = {(5, 1), (5,2), (5, 3), (5, 4), (5, 5) (5, 6)} ⇒ n (F) = 6
⇒ E ∩ F = ((5,5), (5,6)} ⇒ n(E ∩ F) = 2
The conditional probabilty of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P(E/F)
(ii) Let E : set of events having 8 as the sum of the observations,
F : set of events in which red die resulted in a (in any one die) number less than 4
∴ E = {(2, 6), (3, 5), (4, 4), (5,3), (6,2),} ⇒ n (E) = 5
Question 6.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad one will be approved for sale.
Answer:
Let A, B and C be the respective events that the first, second and third drawn orange is good.
Therefore, probability that first drawn oragne is good, P(A) = \(\frac{12}{15}\)
Therefore, probability of getting second orange is good, P(B) = \(\frac{11}{14}\)
(∵ The oranges are not replaced so number of good oranges left is 11)
Similarly, probability of getting third orange is good, P(C) = \(\frac{10}{13}\)
(∵ The oranges are not replaced so number of good oranges left is 10)
The box is approved for sale, if all the three oranges are good.
Thus, probability of getting all the three oranges good = \(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}\)
Therefore, the probability that the box is approved for sale = \(\frac{44}{91}\)
Question 7.
A fair coin and an unbiased die’are tossed. Let A be the event ‘head appears on the coin’ and B be the events ‘3 on the die’. Check whether A and B are independent events or not.
Answer:
If a fair coin and an unbiased die are tossed, then the sample space S is given by,
Also, A: head appears on the coin B : 3 appears on the die.
∴ A = set of events having head on one coin
= {(H, 1), (H, 1), (H, 3), (H, 4), (H, 5), (H, 6)},
B = set of events having 3 on one die = {(H, 3), (T, 3)}
⇒ A ∩ B = {(H,3)} ⇒ n(A) = 6, n(B) = 2, n(A ∩ B) = 1
Therefore, A and are independent events.
Question 8.
A die marked 1,2,3 in red and 4,5,6 in red and 4,5,6 in green is tossed. Let A be the event, ‘number is even’ and B be the event, ‘number is red’. Are A and B independent?
Answer:
When a die is thrown, the sample space S = {1,2, 3,4, 5,6}.
Also, A : number is even and B : number is red
∴ A = {2,4,6}, B = {1,2,3} and A ∩ B = {2}.
⇒ n(A) = 3, n(B) = 3, n(A ∩ B) = 1
Thus, A and B are not independent events.
Question 9.
Let E and F be events with P(E) = \(\frac { 3 }{ 5 }\), P(F) = \(\frac { 3 }{ 10 }\) and P(E ∩ F) = \(\frac { 1 }{ 5 }\). Are E and F independent?
Answer:
Therefore, E and F are not independent events.
Question 10.
Given that the events A and Bare such that P (A) = 1/2, P(A ∪ B) = 3 / 5 and P(B) = p. Find p, if they are
(i) mutually exclusive
(ii) independent
Answer:
(i) When A and B are mutually exclusive, then
A ∩ B = Φ ⇒ P(A ∩ B) = 0
∴ P(A ∪ B) = P(A) + P(B) – PP(A ∩ B) ⇒ \(\frac{3}{5}=\frac{1}{2}\) + p – 0,
(ii) When Aand B are mutually events, then
Now, P(A ∪ B) = P(A).P(B) – P(A) × P(B)
Question 11.
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find.
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(A/B)
(iv) P(B/A)
Answer:
It is given that P(A) = 0.3 and P(B) = 0.4
(i) If A and b are independent events, then
P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12
(ii) We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A)P(B)
= 0.3 + 0.4 – 0.3 × 0.4 = 0.7 – 0.12 = 0.58
Question 12.
If A and B are two events such that P(A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\) and P(A ∩ B) = \(\frac { 1 }{ 8 }\), P(not A and not B).
Answer:
We have, P(A) = \(\frac { 1 }{ 4 }\), P(B) = \(\frac { 1 }{ 2 }\) and P(A ∩ B) = \(\frac { 1 }{ 8 }\)
Therefore, A and B are independent events.
⇒ A’ and B’ are also independent events.
⇒ P(A’ ∩ B’) = P(A’)P(B’)
∴ P (not A and not B) = P(A’ ∩ B’ ) = P(A’) P(B’) = \(\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}\)
Question 13.
Events A and B are such that P(A) = \(\frac { 1 }{ 2 }\), P(B) = \(\frac { 7 }{ 12 }\) P(not A or not B) = \(\frac { 1 }{ 4 }\) State whether A and B are independent?
Answer:
Given, P (not A or not B) = \(\frac { 1 }{ 4 }\) ⇒ P(A’or B’) = \(\frac { 1 }{ 4 }\)
⇒ P(A) × P(B) ≠ P(A ∩ B)
∴ A and B are not independent events.
Question 14.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P (A and B)
(ii) P (A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Answer:
It is given that P(A) = 0.3 and P(B) = 0.6
Also, A and B are independent events.
(i) P (A and B) = P(A ∩ B) = P(A) × P(B) = 0.3 × 0.6 = 0.18 ,
(∵ A and B are independent)
(ii) P (A and B) = P(A ∩ B’) = P(A) × P (B’)
(∵ A and B are independent, ∴ A and B’ are also independent)
= (0.3) [1 – P(B)] = (0.3) (1 – 0.6) = 0.3 × 0.4 = 0.12
(iii) P(A or B)= P(A ∪ B)
= P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) × P(B)
= 0.3 + 0.6 – 0.3 × 0.6 = 0.9 – 0.18 = 0.72
(iv) P (neither A nor B) = P(A’ and B’)
= P(A’ ∩ B’) = P(A’ ∪ B)’
[∵ P(A’ ∩ B’) = P(A’ ∪ B)’]
= 1 – P(A ∪ B)
= 1 – 0.72 = 0.28. (Using parts (iii)]
Question 15.
A die is tossed thrice. Find the probability of getting an odd number least once.
Answer:
When a die is thrown, there are 3 odd numbers on the die out of 6 numbers.
∴ Probability of getting an odd number
Probabiity of getting an even number
= 1 – probability of getting an odd number = \(1-\frac{1}{2}=\frac{1}{2}\)
Probability of getting an even number when the die is tossed thrice
P(E) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}\)
Therefore, probbility of getting an odd number atleast once
= 1 – Probability of getting an odd number in none of the throws
= 1 – P(E) = \(1-\frac{1}{8}=\frac{7}{8}\)
Question 17.
An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?
Answer:
The two balls are selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball and X represents the number of black balls.
X(BB) = 2, X(BR) = 1, X(RB) = 1 and X(RR) = 0
Therefore, the possible values of X are 0, 1 and 2.
Thus, X is a random variable.
Question 18.
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer:
Let X denotes the random variable which denotes the number of tails when a biased coin is tossed twice.
So, X may have value 0, 1 or 2.
Since, the coin is biased in which head is 3 times as likely to occur as a tail.
∴ P{H} = \(\frac { 3 }{ 4 }\) and P{T} = \(\frac { 1 }{ 4 }\)
P(X = 0) = P{HH} = \(\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\)
P (X = 1) = P (one tail and one head)
= P {HT, TH} = P {HT} + P{TH} = P{H} P{H} +P{T} {H}
Question 19.
Two cards are drawn successfully with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of number of aces.
Answer:
Let X denote the number of aces.
X is a random variable.
X can take values 0, 1 or 2.
P(X = 0) = P (non ace and ace)
= P (non ace) × P (ace)
= \(\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)
P(x = 1) = P (ace and non ace or non ace and ace)
= P (ace and non ace) + P (non ace and ace)
= P (ace) × p (non ace) + P (non ace) × P (ace)
∴ Probability distribution of X is
Question 20.
A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?
Answer:
S = {(b,b),(g,b),(b,g),(g,g)}
Let A : both the children are boys
B : at least one of the child is a boy.
A={(b,b)} B = {(b,b),(g,b),(b,g)}
P(B) = § P(AnB) = -t
Question 21.
A fair die is rolled. Consider events E = {1, 3, 5} F = {2, 3} and G = {2, 3, 4, 5}. Find
Answer:
Here, the sample space is S = {1, 2, 3, 4, 5, 6},
Given, E = {1, 3, 5} F = {2,3} and G= {2,3, 4, 5}
⇒ E ∩ F = {3}, E ∩ G = {3,5},
E ∪ F = {1,2,3,5}, (E ∪ F) ∩ G = {2,3,5} and (E ∩ F) ∩ G = {3}.
⇒ n (S) = 6, n(E) = 3, n(F) = 2, n(G) = 4
n(E ∩ F) = 1, n(E ∩ G) = 2, n(E ∩ F) = 4
n[(E ∩ F) ∩ G] = 3 and n[(E ∩ F) ∩ G] = 1
Question 22.
Assume that each child born is equally likely to be boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl?
(ii) atleast one is a girls?
Answer:
Let b and g represent the boy and the girl child, respectively.
If a family has two children, the sample space will be
S = {bb, bg, gb, gg}
which contains four equally likely sample points i.e., n (S) = 4
Let E : both children are girls, then E = {gg} ⇒ n (F) = 1
(i) Let F : the youngest is a girl, then F = {bg, gg} ⇒ n (F) = 2
⇒ E ∩ F = {gg} ⇒ n(E ∩ F) = 1
(ii) Let F : atleast one is a girl, then F = {bg,gb,gg} EF= {gg} = E ∩ F = {gg} = E
⇒ n(F) = 3, n(E ∩ F) = 1
Question 23.
An instructor has a question bank consisting of 300 easy true/false questions, 200 difficult true/false question, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer:
Total number of questions = 300 + 200 + 500 + 400 = 1400.
Let E be the event that selected question is an easy question
Then, n(E) = 500 + 300 = 800
Let F be the event that selected question is an easy question
Then, n(F) = 500 + 400 = 9
Question 24.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the events ‘the sum of numbers on the dice is 4’.
Answer:
When dice is thrown, number of observations in the sample space S = 6 × 6 = 36 (equally likely sample events) i.e., n (S) 36
Let E : set of numbers in which numbers appearing on the two dice re different
Then, E = {(1, 3), (2, 2), (3. l)} ⇒ n(E) = 3
n(F) = 30
Here, F contains all points of S except {(1, 1), (2, 2), (3, 3),(4, 4), (5, 5), (6, 6))
∴ E ∩ F = {1,3),(3,1)}
Question 25.
Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event the coin shows a tail, given that atleast one die shows a 3.
Answer:
The outcomes of the given experiment can be represented by the following set.
The sample space of the experiment is,
n = (S) = 20
Let E : the coin shows a tail, F : atleast one die shows up a 3,
E = 4(1, T), (2, T), (4, l’),(5, T)}, ⇒ n (E) = 4
E = {(3, 1), (3, 2), (3, 3), (3,4), (3, 5), (3, 6),(6, 3)}, ⇒ n (F) = 7
⇒ E ∩ F = Φ because here is no common elements.
Question 26.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a balls is drawn at random. What is the probability that the second ball is red?
Answer:
The urn contains 5 red and 5 black balls.
i.e.,n (R) = 5, n (B) = 5 and n (S) = 10
Let a red ball be drawn in the first attempt
∴ P(drawing a red ball) = \(\frac{n(R)}{n(S)}=\frac{5}{10}=\frac{1}{2}\)
If two red balls are added to the urn, then the urn contains 7 red and
5 black balls i.e., n(R) = 7, n(B) = 5 and n(S) = 12
P(drawing a red ball) = \(\frac{n(R)}{n(S)}=\frac{7}{12}\)
Let a black ball be drawn in the first attempt
Then, n (R) = 5, n (B) = 5 and n(S) = 10
P (drawing a black ball in the first attempt) = \(\frac{n(B)}{n(S)}=\frac{5}{10}=\frac{1}{2}\)
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
i.e., n(R) = 5 n(B) = 7 and n(S) = 12 ∴ P(drawing a red ball) = \(\frac{n(R)}{n(S)}=\frac{5}{12}\)
Therefore, probability of drawing second ball as red is
Question 27
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer:
Let E1 : first bag is selected, E2 : second bag is selected
Then, E1 and E2 are mutually exclusive and exhaustive. Moreover,
P(E1) = P(E2) = \(\frac { 1 }{ 2 }\)
Let E : ball drawn is red.
Question 28.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tosses 6 times. What are possible values of X?
Answer:
Let M denotes the number of heads and N denotes the number of ails when a coin is tossed 6 times. Then,
Given, X = difference between M and N = |M – N|
Here, both M and N can take values 0, 1,2, 3, 4, 5, 6 but M + N is always equal to 6. Look at the following table:
Thus, we see that X takes values, 0, 2, 4 and 6.
Question 29.
Find the probability distribution of
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Answer:
(i) When one coin is tossed twice, the sample space is
S = {HH, HT, TH, TT}.
Let X denotes, the number of heads in any outcome in S,
X(HH) = 2, X (HT) = 1, X(TH) = 1 and X(TT) = 0 .
Therefore, X can take the value of 0, 1 or 2. It is known that
P(HH) = P(HT) = P(TH) = P(TT) = \(\frac { 1 }{ 4 }\)
∴ P(X = 0) = P (tail occurs on both tosses) = P({TT}) = \(\frac { 1 }{ 4 }\)
P (X = 1) = P (one head and one tail occurs) = P({TH,HT}) = \(\frac{2}{4}=\frac{1}{2}\)
and P(X = 2) = P (head occurs on both tosses) = P({HH}) = \(\frac { 1 }{ 4 }\)
Thus, the required probability distribution is as follows.
(ii) When three coins are tossed thrice, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
which contains eight equally likely sample points.
Let X represent the number of tails.
Then, X can take values 0, 1, 2 and 3.
P (X = 0) = P (no tail) = P ({HHH}) = \(\frac { 1 }{ 8 }\)
P (X = 1) = P (one tail and two heads show up)
= P ({HHT, HTH, THH}) = \(\frac { 3 }{ 8 }\),
P (X = 2) = P (two tails and one head show up)
= P ({HTT, THT, TTH}) = \(\frac { 3 }{ 8 }\)
and P (X = 3) = P (three tails show up) = P ({TTT}) = \(\frac { 1 }{ 8 }\).
Thus, the probability distribution is as follows
(iii) When a coins is tossed four times, the sample space is
which contains 16 equally likely sample points.
Let X be the random variable, which represent the number of tails, heads. It can be seen that X can take the value of 0, 1, 2, 3 or 4.
P (X = 0) = P (no head shows up) = P ({TTTT}) = \(\frac { 1 }{ 16 }\),
P (X = 1) = P (one head and three tails show up)
= P ( {HTTT, THTT, TTHT, TTTH} ) = \(\frac{4}{16}=\frac{1}{4}\)
P (X = 2) = P (two heads and two tail show up)
= P ({HHTT, HTHT, HTTH, THHT, THTH, TTHH}) = \(\frac{6}{16}=\frac{3}{8}\)
P (X = 3) = P (three heads and one tail show up) .
= P ({HHHT, HHTH, HTHH, THHH}) = \(\frac{4}{16}=\frac{1}{4}\)
and P (X = 4) = P (four heads show up) = P ({HHHH}) = \(\frac{1}{16}\)
Thus, the probability distribution is as follows
Question 30.
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer:
Let X denotes the random variable which denotes the number of tails when a biased coin is tossed twice.
So, X may have value 0, 1 or 2.
Since, the coin is biased in which head is 3 times as likely to occur as a tail.
∴ P{H} = \(\frac { 3 }{ 4}\) and P{T} = \(\frac { 1 }{ 4 }\)
P(X = O) = P{HH} = \(\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\)
P (X = 1) = P (one tail and one head)
= P {HT, TH} = P {HT} + P(TH} = P{H} P{H} + P{T} {H}
P(X = 2) = P (two tails) = P{TT} = P{T}. P{T} = \(\left(\frac{1}{4}\right)^{2}=\frac{1}{16}\)
Therefore, the required probability distribution is as follows.
2nd PUC Maths Probability Three Marks Questions and Answers
Question 1.
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous years results report thit 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostler?
Answer:
Let E1 : the event that the student is residing in hostel and E2 : the event that the student is not
residing in the hosteL
Let E : a student attains A grade,
Then, E1 and E2 are mutually exclusive and exclusive and exhaustive. Moreover,
By using Baye’s theorem, we obtain
Question 2.
In answering a question on a multiple choice test a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4 What is the probability that a student knows the answer given that he answered it correctly?
Answer:
Let E1 : the event that the studennt knows the answer and E2 : the event that the student guesses the answer.
Therefore, E1 and E2 are mutually exclusive and exhaustive.
∴ P(E1) = \(\frac { 3 }{ 4 }\) and P(E2) = \(\frac { 1 }{ 4 }\)
Let E : the answer is correct.
The probability that the student answered correctly, given that he knows the answer, is 1 i.e. P(E/E1) = 1
Probability that the student answered correctly, given that he guessed, is \(\frac { 1 }{ 4 }\) i.e., P(E/E2) = \(\frac { 1 }{ 4 }\)
By using Baye’s theorem, we obtain
Question 3.
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested, then with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually lias the disease, what is the probability that a person has disease given that his test result is positive?
Answer:
Let E1 : the event that the person has disease and E2 : the events that the person is healthy.
Then, E1 and E2 are mutually exclusive and exhaustive.
Moreover, P(E1) = 0.1% = \(\frac{0.1}{100}\) = 0.001 and P(E2) = 1 – 0.001 = 0.999
Let E : the event that test is positive,
P(E/E1) = P (result is positive given the person has disease)
99% = \(\frac{99}{100}\) = 0.99
Probability that a person does not have disease and test result is positive.
By using Baye’s theorem, we obtain
Question 4.
There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows head, what is the probability that it is was the two headed coin?
Answer:
Let E1 : the event that the coin chosen is two headed,
E2 : the event that the coin chosen is biased
and E3 : the event that the coin chosen is unbiased
⇒ E1, E2, E3 are mutually exclusive and exhaustive events. Moreover,
P(E1) = P(E2) = P(E3) = \(\frac { 1 }{ 3 }\)
Let E : tosses coin shows up a head,
∴ P(E/E1) = P (coin showing heads, given that it is a two headed coin) = 1
P(E/E2) = P (coin showing heads, given that it is a biased coin)
75% = \(\frac{75}{100}=\frac{3}{4}\)
P(E/E3) = P (coin showing heads, given that it is an unbiased coin) = \(\frac { 1 }{ 2 }\)
The probability that the coin is two headed, gi’en that it shows head, is given by P(E/E1)
By using Baye’s theorem, we obtain
Question 5.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 ^respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
There are 2000 scooter drivers, 4000 car drivers and 6000 truck drivers.
Total number of drivers = 2000 + 4000 + 6000 =12000
Let E1 : the event that insured person is a scooter driver, E2: the event that insured person is a car driver
and E3 : the event that insured person is a truck driver,
Then, E1, E2, E3 are mutually exclusive and exhaustive events. Moreover,
Let E : the event that insured person meets with an accident,
P(E/E1) = P (scooter driver met with an accident) = 0.01 = \(\frac { 1 }{ 100 }\)
P(E/E2)= P (car driver met with an accident) = 0.03 = \(\frac { 3 }{ 100 }\)
P(E/E3) = P (truck driver met with an accident) = 0.15 = \(\frac { 15 }{ 100 }\)
The probability that the driver is a scooter driver, given he met with an accident, is given by P(E1/E)
By using Baye’s theorem, we obtain
Question 6.
A factory has two machines A and B. Past record shows that machine A produced 60% of the item of output and machine B produced 40% of the items. Further, 2% of the items produce by machine A and 1% produced by machine B were efective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer:
Let E1 : the event that the item is produced by machine A and E2 : the event that the item is produced by machine B.
Then, E1 and E2 are mutually exclusive and exhaustive events. Moreover,
Let E : the event that the item chosen is defective,
∴ P(E/E1) = P (Machine A produced defective items) = 2% = \(\frac { 2 }{ 100 }\)
P(E/E2) = P (Machine B produced defective items) = 1% = \(\frac { 1 }{ 100 }\)
The probability that the randomly selected item was from machine B, given that it is detective, P(E2/E)
By using Baye’s theorem, we obtain
Question 7.
Two groups are competing for the position on the board of directors of a corporation. The probability that the first and the second groups will win are 0.6 and 0.4, respectively. Further, if the first group wins the probability of introducing anew product is 0.7 and the corresponding probability is 0.3 , if the second group wins. Find the probability that the new product introduce was by the second group.
Answer:
Let E1 : the event that the first group wins and E2 : the event that the second group wins.
Then, E1 and E2 are mutually exclusive and exhaustive events.
∴ P(E1) = 0.6 = \(\frac { 6 }{ 10 }\) and P(E2) = 0.4 = \(\frac { 4 }{ 10 }\)
Let E : the event that the new product is introduced.
P(E/E1) = P (introducing a new product, if the first group wins) = 0.7 = \(\frac { 7 }{ 10 }\)
∴ P(E/E2) = P (introducing a new product, if the second group wins) = 0.3 = \(\frac { 3 }{ 10 }\)
The probability that the new product is introduced by the second group is given by p\(\left(\frac{E_{2}}{E}\right)\)
By using Baye’s theorem, we obtain
Question 8.
Suppose, a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and * notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once an notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer:
Let E1 : the event that 5 or 6 is shown on die
and E2 : the event that 1, 2, 3, or 4 is shown on die.
Then, E1 and E2 are mutually exclusive and exhaustive events,
and n (E1) = 2, n (E2) = 4
Also, n (S) = 6
Let E : the event that exactly one head show up,
∴ P(E/E1) = P (exactly one head show up when coin is tossed thrice)
= P{HTT,THT,TTH} = \(\frac { 3 }{ 8 }\) (∵ total number of events = 23 = 8)
P(E/E2) = P (head shows up when coin is tossed once) = \(\frac { 1 }{ 2 }\)
The probability that the girl threw, 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by P(E2/E)
By using Baye’s theorem, we obtain
Question 9.
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items, respectively. A is on the job for 50% of the time, B on the fob for 30% of the time and C on the fob for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Answer:
Let E1 : the event that item is produced by machine A.
E2 : the event that item is produced by machine B
and E3 : the event that item is produced by machine C
Here, E1, E2 and E3 are mutually exclusive and exhaustive event Moreover,
Let E : The event that item chosen is found to be defective’,
By using Baye’s theorem, we obtain
Question 10.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be diamonds. Find tbe probability of the lost card being a diamond.
Answer:
Let E1 : the event that lost cards is a diamond ⇒ n(E1) = 13
E2 : lost cards is not a diamond ⇒ n (E2) = 52 – 13 = 39
And, n (S) = 52
Then, E1 and E2 are mutually exclusive and exhaustive events.
Let E : the events that two cards drawn from the remaining pack are diamonds,
When one diamond card is lost, there are 12 diamond cards out of 51 cards.
The cards can be drawn out of 12 diamond cards in 12C2 ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in 51C2 ways. The probability of getting two cards, when one diamond card is lost, is given by P(E/E1)
2nd PUC Maths Probability Five marks Questions and Answers
Problems and Solutions on Bernoulli trials and binomial distribution.
Question 1.
From a lot of 30 bulbs which include 6 defective, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer:
It is given that out of 30 bulbs, 6 are defective.
Number of non-defective bulbs = 30 – 6 = 24
4 bulbs are drawn from the lot with replacement.
Let p = P(obtaining a defective bulb when a bulb is drawn) = \(\frac{6}{30}=\frac{1}{5}\)
q = P(obtaining a non-defective bulb when a bulb is drawn) = \(\frac{24}{60}=\frac{4}{5}\)
P(X = 0) = P (no defective bulb in the sample)
Therefore, the required probability distribution is as follows.
Question 2.
A die is thrown 6 times. If getting an old number is a success what is the probability of (i) 5 success (ii) at least 5 success (iii) at most 5 success.
Answer:
Let P be the probability of success (getting an odd number)
∴ p = \(\frac{3}{6}=\frac{1}{2}\)
p + q = 1 ⇒ q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\)
n = 6
Let X denote the number of success in the experiment.
The possible values of X are 0, 1,2, 3, 4, 5, 6 ,
P (X = r) = nCrprqr-1
Question 3.
A fair coin is tossed 10 times. Find the probability of
(i) exactly 6 heads
(ii) at least 6 heads
(iii) at most 6 heads.
Answer:
Let X denote the number of heads in an experiment of 10 trials.
(ii) P (at least 6 heads) = P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
(iii) P (at most 6 heads) =
P(X ≤ 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
Question 4.
A pair of dice is thrown 4 times. If getting a doublet is considered a success find the probability of 2 success.
Answer:
Let P be the probability of success of getting a doublet j4-OOj
{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
Let X be the number of success.
n = 4 p = \(\frac { 1 }{ 6 }\) q = \(\frac { 5 }{ 6 }\)
P(X = r) = nCrprqr-1
Question 5.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than 1 defective item.
Answer:
Question 6.
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
(i) all the five cards are spades
(ii) only 3 cards are spades
(iii) one is a spade
Answer:
Question 7.
A person buys a lottery ticket in 5 lotteries in each of which his chance of winning a prize is \(\frac { 1 }{ 100 }\). What is the probability that he will win a prize
(a) at least once
(b) exactly once
(c) at least twice?
Answer:
Let X denote the number of wins
Question 8.
Find the probability of getting 5 exactly twice in 7 throws of a die.
Answer:
probability of getting a 5 in throwing a die = \(\frac { 1 }{ 6 }\)
Question 9.
On a multiple choice questions with three possible answers for each of the five questions, what is the probability that a candidate would get 4 or more correct answers just by guessing?
Answer:
Let X denote the number of correct answers p = \(\frac { 1 }{ 3 }\) q = 1 – \(\frac{1}{3}=\frac{2}{3}\)
