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If the family has 3 children, then they can be

{BBB, BBG, BGB, GBB, BGG, GBG, GBB, GGG}

total number of combinations = 8

number of combinations with atleast one boy = 7

Probability(atleast one boy) =

{BBB, BBG, BGB, GBB, BGG, GBG, GBB, GGG}

total number of combinations = 8

number of combinations with atleast one boy = 7

Probability(atleast one boy) =

**7/8**
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Some times it is simpler to calculate the contrary of the event and its probability and then subtract it from 1 to get our answer.

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The probability of one child being a boy = 1/2 a girl = 1/2

Let us assume that the 3 children being boys or girls are independent events.

Event of having no boys at all

= first child is a girl

P(no boys among 3 children) = 1/2 * 1/2 * 1/2 = 1/8

P(there is at least one boy among 3 children) = 1 - 1/8 = 7/8

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The probability of one child being a boy = 1/2 a girl = 1/2

Let us assume that the 3 children being boys or girls are independent events.

Event of having no boys at all

= first child is a girl

**2nd child is a girl***AND***3rd girl is a girl.***AND*P(no boys among 3 children) = 1/2 * 1/2 * 1/2 = 1/8

P(there is at least one boy among 3 children) = 1 - 1/8 = 7/8