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Home  >>  CBSE XII  >>  Math  >>  Probability
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Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

$\begin{array}{1 1} \text{P (youngest is a girl) = 1/2, P (at least one is a girl) = 1/3 }\\\text{P (youngest is a girl) = 1/3, P (at least one is a girl) = 1/2} \\\text{P (youngest is a girl) = 1/3, P (at least one is a girl) = 1/4} \\ \text{P (youngest is a girl) = 1/2, P (at least one is a girl) = 2/3} \end{array} $

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1 Answer

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Toolbox:
  • In any problem that involves determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
  • Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
If a family has 2 children, the sample space can be reprsented as S = {(b,b), (b,g), (g,b), (g,g)}. Thus the total number of outcomes is 4.
The total number of outcomes for the event that both children are girls is 1 which is E = {g,g} $\rightarrow P(E) = \large \frac{1}{4}$.
Case a) If the youngest is a girl, then the total number of favourable outcomes for this event is 2, which are F = {(b,g), (g,g)} $\rightarrow P(F) = \large \frac{2}{4} = \frac{1}{2}$.
$\Rightarrow E \cap F = {g,g} \rightarrow P (E \cap F) = \large \frac{1}{4}$
Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
$\Rightarrow P(E/F) = \Large \frac{\frac{1}{4}}{\frac{1}{2}}$ = $\large \frac{1}{2}$
Case b) If atleast one of the children is a girl, then the total number of favorable outcomes is 3, F = {(b,g), (g,b), (g,g)} $\rightarrow P(E) = \large \frac{3}{4}$
$\Rightarrow E \cap F = {g,g} \rightarrow P (E \cap F) = \large \frac{1}{4}$
$\Rightarrow P(E/F) = \Large \frac{\frac{1}{4}}{\frac{3}{4}}$ = $\large \frac{1}{3}$
answered Mar 5, 2013 by poojasapani_1
edited Jun 18, 2013 by balaji.thirumalai
 

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