6. In a family of two children , a child is selected at random . Find the
probability of
(i) Both are boys.
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6. In a family of two children , a child is selected at random . Find the
probability of
(i) Both are boys.
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Answer:
find P (both children are males, if it is known that at least one of the children is male).
A: Event that both children are male, and B: event that at least one of them is a male.
A: {MM} and B: {MF, FM, MM} →→ P (A ∩∩ B) = {MM}
Probability that both are males, if we know one is a female =P(A/B)=P(A∩B)/P(B)
(P(A∩B)P(B))
Given S = {MM, MF, FM, FF}, we can see that: P (A) = 1414; P (B) = 3434; P (A ∩∩ B) = 1414
Therefore P(B/A)=P(A∩B)/P(A) =(1/4)/(3/4)=1/3
am starting with the assumption that we are selecting from the population of ALL families, not just the ones that have boys.
In the first method, I am assuming that we are randomly choosing one child from the family that we selected, and observing that child, but not the other.
The possibilities are:
BB
BG
GB
GG
When we observe a child at random, each of these eight children has an equal chance of being the one
we saw.
We make our observation, and the child we see is a boy.
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