A father has three children with at least one boy. The probability that he has 2 boys and 1 girl is

1/4

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1/3

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2/3

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None of these

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Solution

The correct option is A 1/3
Let's represent boy with B and girl with G
Then possible outcomes are BBB,BBG,BGG
And the favorable outcome is BBG
Let's call this event to be E
Then
$\Rightarrow P (E) = \frac{n o . o f f a v o r a b l e o u t c o m e s}{n o . o f p o s s i b l e o u t c o m e s} \Rightarrow P (E) = \frac{1}{3}$

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A father has three children with at least one boy. The probability that he has 2 boys and 1 girl is

The correct option is A 1/3Let's represent boy with B and girl with GThen possible outcomes are BBB,BBG,BGGAnd the favorable outcome is BBGLet's call this event to be EThen⇒P(E)=no.of favorable outcomesno.of possible outcomes⇒P(E)=13