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Question
2.
Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
Solution :
Given: P (B) = 0.5, P (A
∩
B) = 0.32
(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ E ∩ F = {HHT, HTH, THH}
(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT, TTH, TTT}
Question
7. Determine :Two coins are tossed once.
(i) E : tail appears on one coin, F : one coin shows head.
(ii) E : no tail appears, F : no head appears.
Solution :
If two coins are tossed once, then the sample space S is
S = {HH, HT, TH, TT}
(i) E = {HT, TH}
F = {HT, TH}
(ii) E = {HH}
F = {TT}
∴ E ∩ F = Φ
P (F) = 1 and P (E ∩ F) = 0
∴ P(E|F) =
Question
8. Determine :E A dice is thrown three times.
E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
Solution :
Since a dice has six faces. Therefore E = 6 x 6 x 6 = 216
E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)
F = (6) x (5) x (1, 2, 3, 4, 5, 6)
Question
9. Determine : Mother, father and son line up at random for a family picture.
E : Son on one end, F : Father in middle.
Solution :
If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
∴ E ∩ F = {MFS, SFM}
E : Son on one end
Question
10. A black and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution :
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 has 6 × 6 = 36 number of elements.
(a) Let A: Obtaining a sum greater than 9
= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5.
= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).
(b) E: Sum of the observations is 8.
= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F: Red die resulted in a number less than 4.
The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).
Question
11. A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find:
(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}
E ∩ F = {3}
(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}
Question
12. Assume that each child born is equally likely to be a 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) at least one is a girl?
Solution :
Let
b
and
g
represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(
b
,
b
), (
b
,
g
), (
g
,
b
), (
g,
g)}
Let A be the event that both children are girls.
∴ A = {(g,g)}
(i) Let B be the event that the youngest child is a girl.
The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).
Therefore, the required probability is 1/2.
(ii) Let C be the event that at least one child is a girl.
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).
Question
13. An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 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?
Solution :
The given data can be tabulated as
P(E|M) = P(E ∩ M)/P(M) = 5/14/9/14 = 5/9
Therefore, the required probability is 5/9.
Question
14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution :
When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.
∴ A = {(1, 3), (2, 2), (3, 1)
S = (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)
(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)
B = {(2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2),
(3, 2), (4, 2), (5, 2), (6, 2),(1, 3), (2, 3), (4, 3),
(5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4),
(1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (1, 6), (2, 6),
(3, 6), (4, 6), (5, 6)}
Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.
Therefore, the required probability is 1/15.
Question
15. 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 “at least one die shows a 3”.
Solution :
The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
(1, H), (2, H), (3, H), (4, H), (5, H), (1, T), (2, T), (3, T), (4, T), (5, T)}
Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.
Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).
Therefore,
Question
16. If P (A) = 1/2 P (B) = 0, then P(A|B) is:
(A) 0
(B) 1/2
(C) not defined
(D) 1
Solution :
P (A) = 1/2 P(B) = 0
Therefore, P (A|B) is not defined.
Therefore, option (C) is correct.
Question
17.
If A and B are events such that P (A|B) = P(B|A), then
⇒ P (A) = P (B)
Therefore, option (D) is correct.
