suppose a family has three children.find the probability that the first two children born are boys. What is the probability that the last two children are girls?
i cna solve this problem. but, i m not sure if my answer is right or wrong.person who makes the most easiest way is going to get 10marks.
2006-09-19 22:24:40 · 5 answers · asked by free aung san su kyi forthwith 2 in Science & Mathematics ➔ Mathematics
Only 8 (2^3) possibilities exist:
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
let X=don't care,
BBX occurs only twice, so P(BBX) = 2/8 =0.25
XGG occurs only twice, so P(XGG) = 2/8 = 0.25
For larger fields, this brute force method becomes impossible, so
p(B) = 0.5
p(G) = 0.5
p(X)==1
P(BBX) = P(B)*P(B)*P(X) = 0.5*0.5*1 = 0.25
P(XGG) = P(X)*P(G)*P(G) = 1*0.5*0.5 = 0.25
2006-09-19 23:24:56 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
The probability that the first child is a boy is 50%.
The probability that the second child is a boy is 50% again, no matter whether the first child is a boy or not. So the probability that the first two children are both boys is 25%, no matter which children (if any) come along later.
So if there are eight families with two children each, only two of them have two boys, and if they have a third child, only one of them will be boy-boy-girl, and the other will be three boys. It's not completely clear from your question whether you want just the probability that the first two are boys, not caring about the third (that's 25%), or two boys then a girl (that's 12.5%).
The last two being girls is exactly the same - 25% if the first doesn't matter, 12.5% if the first has to be a boy.
2006-09-20 05:50:13 · answer #2 · answered by bh8153 7 · 0⤊ 0⤋
The probability of two boys is 1/2* 1/2 = 1/4. As the birth of boy and girl are equally likely and mutually exhaustive, there is no need to consider the third kid.
Similarly, the P of last two kids to be girls is 1/4
2006-09-20 05:44:57 · answer #3 · answered by Amit K 2 · 1⤊ 0⤋
assuming an equal probability of having a boy or a girl.
answer 1 = probability of first having two boys and a girl + probability
of having three boys
= .5*.5*.5 + .5*.5*.5
answer 2 = probability of first having a boy and two girls + probability
of having three girls
= .5*.5*.5 + .5*.5*.5
2006-09-20 09:06:14 · answer #4 · answered by tronic_hobbist 2 · 0⤊ 0⤋
Hmm... I would have to say... zero... since you stated that the family only has 3 children and the first two are boys... then the probability of the next two being girls is zero since they only have 3 children!
2006-09-20 05:35:38 · answer #5 · answered by Laurie V 4 · 0⤊ 2⤋