What is the probability that a family with two children has two boys given that they have at least one boy *?

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London University

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Hint: To solve this question, we will use the conditional probability. First, we will find the sample set and then we will find the event set that one of the two children are boys and an event set that both the children are boys. Then we will find the event set that both the aforementioned events are happening. Then, we will use the relation of conditional probability that one event is happening, given that another has already happened. This relation is given as P(\[\left. \text{B} \right|\text{A}\]) = $ \dfrac{\text{P}(A\cap B)}{\text{P}(\text{A})} $ , where A and B are two independent events, $ A\cap B $ is the event of A and B both happening and P(\[\left. \text{B} \right|\text{A}\]) is the probability that B is happening, given that event A has happened.

Complete step-by-step answer:

Let b denote a boy and g denote a girl.Let S be the event that a family has two children.Thus, S = {(b, g), (g, b), (g, g), (b, b)}And the number of elements in event set S = n(S) = 4Let A be the event that at least one of the children is a boy.Thus, A = {(b, g), (g, b), (b, b)}And the number of elements in event set A = n(A) = 3Let B be the event that both the children are boys.Thus, B = {(b, b)}The probability that one of the children is boy is given as P(A) = $ \dfrac{\text{n}\left( \text{A} \right)}{\text{n}\left( \text{S} \right)} $ . $ \Rightarrow $ P(A) = $ \dfrac{3}{4} $ The probability that both the children are boys is given as P(B) = $ \dfrac{\text{n}\left( \text{B} \right)}{\text{n}\left( \text{S} \right)} $ . $ \Rightarrow $ P(B) = $ \dfrac{1}{4} $ \[\text{A}\cap \text{B}\] = {(b, b)}The number of elements in the event set \[\text{A}\cap \text{B}\] = n(\[\text{A}\cap \text{B}\]) = 1.The probability event A happening and event B happening is given as P(\[\text{A}\cap \text{B}\]) = $ \dfrac{\text{n}\left( \text{A}\cap \text{B} \right)}{\text{n}\left( \text{S} \right)} $ $ \Rightarrow $ P(\[\text{A}\cap \text{B}\]) = $ \dfrac{1}{4} $ Now, from the concepts of conditional probability, we know that the probability of B given that A is happening is given by P(\[\left. \text{B} \right|\text{A}\]) = $ \dfrac{\text{P}(A\cap B)}{\text{P}(\text{A})} $, where P(\[\text{A}\cap \text{B}\]) is the probability of A and B both happening and P(A) is the probability of A happening. $ \Rightarrow $ P(\[\left. \text{B} \right|\text{A}\]) = $ \dfrac{\dfrac{1}{4}}{\dfrac{3}{4}} $ $ \Rightarrow $ P(\[\left. \text{B} \right|\text{A}\]) = $ \dfrac{1}{3} $ Therefore, the probability that both the children are boys given that one of them is boy is $ \dfrac{1}{3} $ .

Note: It is to be noted that the relation P(\[\left. \text{B} \right|\text{A}\]) = $ \dfrac{\text{P}(A\cap B)}{\text{P}(\text{A})} $ can only be used when the two events A and B are independent event as given in our question. The event of the 1st boy or girl does not affect the gender of the second child. If the events A and B depend on each other, students have to use Bayes’ theorem.

Last updated at May 25, 2021 by Teachoo

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Example 2 A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ? A Family has two children Let girls be denoted by g & boys be denoted by b S = {(g, g), (g, b), (b, g), (b, b)} We need to find the probability that both the children are boys, given at least one them is a boys F : at least one of the children is a boys E : both the children are boys We need to find P(E|F) Also, E F = {(b, b)} P(E F ) = 1 4 Now, P(E|F) = ( ) ( ) = 1 4 3 4 = 1 3 P(E|F) =

I can understand your confusion. Always write down the total sample space and sample space under condition. This is a question of conditional probability. The total sample space in this case is $\lbrace B, B \rbrace$, $\lbrace B, G \rbrace$, $\lbrace G, B \rbrace$, $\lbrace G, G \rbrace$ each case with equal probability $1/4$. When it says that there is "atleast one boy". Then your conditional sample space shrinks to $\lbrace B, B \rbrace$, $\lbrace B, G \rbrace$, $\lbrace G, B \rbrace$,. Now is asking the probability of having 2 boys, hence your favourable sample space is $\lbrace B, B \rbrace$. Hence the probability is = number of favourable sample space/ number of conditional sample space = 1/3.