F5 math's Q (probability)

2013-03-04 2:29 am

1. Mr and Mrs Lee want to have at least a son and a daughter. How many
children should Mrs Lee give birth to so that the probability of having at least a
son and a daughter is more than 0.9? (ans:5)

2. Tom and Peter are students of S5A. The probability that Tom is absent on a
school day is 0.05 while the probability that Peter is absent on a school day is 0.06. What are the probabilities that

(a) either or both of them is absent, (ans: 0.107)

(b) at least one of them comes to the class, (ans:0.997)

(c) only one of them comes to the class (ans:0.104)

on a certain school day?

可否解釋一下
Thank you very much!

回答 (1)

micatkie

2013-03-04 3:05 am

✔ 最佳答案

1.
Let n be the required number of children.

P(at least a son and a daughter) > 0.9
1 - P(n sons) - P(n daughters) > 0.9
1 - (1/2)ⁿ - (1/2)ⁿ > 0.9
2 x (1/2)ⁿ < 0.1
(1/2)ⁿ < 0.05
n log(1/2) > log(0.05) ...... [for log(0.05) < 0]
n > log(0.05) / log(1/2)
n > 4.32

Hence, the required number of children = 5

=====
2.
(a)
T : Tom is present ...... T' : Tom is absent
P : Peter is present ...... P' : Peter is absent

P(T') = 0.05 and P(T) = 1 - 0.05 = 0.95
P(P') = 0.06 and P(P) = 1 - 0.06 = 0.94

P(either or both of is absent)
= 1 - P(both of them are present)
= 1 - P(T) x P(P)
= 1 - 0.95 x 0.94
= 1 - 0.893
= 0.107

(b)
P(at least one of them comes to the class)
= 1 - P(both of them are absent)
= 1 - P(T') x P(P')
= 1 - 0.05 x 0.06
= 1 - 0.003
= 0.997

(c)
P(only one of them comes to the class)
= P(T and P') + P(T' and P)
= P(T) x P(P') + P(T') x P(P)
= 0.05 x 0.94 + 0.06 x 0.95
= 0.047 + 0.057
= 0.104

參考: micatkie

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