A family has 2 children. The probability that both of them are boys if...
Explanation:
To solve the problem, we need to use conditional probability. Let's consider the following possibilities for the gender of the two children:
- BB (both boys)
- BG (one boy, one girl)
- GB (one girl, one boy)
- GG (both girls)
We know that one of the children is a boy, which eliminates the possibility of GG. So, we are left with three possibilities:
- BB
- BG
- GB
The probability of having two boys (BB) is 1/4, since there are four equally likely possibilities. However, we know that we don't have GG, so the probability of having two boys given that one of them is a boy is:
P(BB | one of them is a boy) = P(BB) / (P(BB) + P(BG) + P(GB))
= 1/4 / (1/4 + 1/2 + 1/2)
= 1/3
Therefore, the probability that both of them are boys if it is known that one of them is a boy is 1/3 or 0.33, which is closest to option B, 1/2.
Conclusion:
The correct option is B, which states that the probability is 1/2. However, this is a common misunderstanding of conditional probability. The correct probability is 1/3, as explained above.
A family has 2 children. The probability that both of them are boys if...
Bcz in family 2 children,
and 1 boy's so probability are 1/2
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