Question

A person has 3 children with at least one boy.
Find the probablitiy of having atleast 2 boys
among the children ?

Answer 1

=]

Answer 2

P(at least 2 boys|1boy)

Answer 3

\(\large \color{black}{\begin{align}
& \normalsize \text{A person has 3 children with at least one boy.} \hspace{.33em}\\~\\
& \normalsize \text{Find the probablitiy of having atleast 2 boys } \hspace{.33em}\\~\\
& \normalsize \text{among the children ?} \hspace{.33em}\\~\\
\end{align}}\)

Answer 4

how to solve this P(at least 2 boys|1boy) ?

Answer 5

you can click edit and put it in the blue part

Answer 6

thnks for suggestion

Answer 7

P(at least 2 boys | at least 1boy) = P(at least 2 boys and at least 1 boy)/P(at least one boy)

P(at least 2 boys)/P(at least one boy)

Answer 8

is this like bayes therem

Answer 9

no just conditional probability

Bayes is a little different

Answer 10

P(at least 2 boys)/P(at least one boy)=(1/3+1/3)/(3/3)

Answer 11

not quite...

P(At least 2 boys) = P(2 boys) + P(3 boys)

P(at least 1 boy) = 1-P(no boys)

Answer 12

=(1/3+1/3)/(1-0/3) =?

Answer 13

P(no boys) = P(all girls) how many children does the couple have? 3.

=> P(all girls) = P(GGG) = (1/2)(1/2)(1/2) = 1/8

Answer 14

final answer=(2/3)/(1/8)=1/12

Answer 15

the area of the smallest circle divded by the bigger circle

Answer 16

no
P(At least 2 boys) = P(2 boys) + P(3 boys)

P(3 boys) = P(3 girls) = 1/8

P(2 boys) = 3C2 * (1/8) = 3/8

P(At least 2 boys) = 1/2

Answer 17

Prob of atleast 2 boys = prob of atleast 1 boy - prob of only 1 boy

prob of @2boys/prob of @ 1 boy = (prob of atleast 1 boy - prob of only 1 boy)/prob of @ 1 boy

= 1- prob_of_1_boy/prob of @ 1 boy

= 1- prob of 1 boy/(1-prob of no boys)

Answer 18

P(at least 2 boys)/P(at least one boy)=(1/2)/(1/8)=1/4 , is it