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  • Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exa

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die ?

 

 

 

 
 
 
 
 

Answers (1)

Let E1 be the even that girl gets 1 or 2 on the roll and
E2 = event that girl gets 3,4,5 or 6 on the roll of a die
\therefore p\left ( E_{1} \right )= \frac{2}{6}= \frac{1}{3}
    \therefore p\left ( E_{2} \right )= \frac{4}{6}= \frac{2}{3}
Let A be the event that she gets exactly one tail . If she tossed coin 3 times and gets exactly one tail then possible outcomes are HTH, HHT, THH
\therefore p\left ( \frac{A}{E_{1}} \right )=\frac{3}{8}
If she tossed coin only once and exactly one tail shows
then p\left ( \frac{A}{E_{2}} \right )=\frac{1}{2}
\therefore p\left ( \frac{E_{2}}{A} \right )=\frac{p\left ( E_{2} \right )\cdot P\left ( \frac{A}{E_{2}} \right )}{p\left ( E_{1} \right )\cdot p\left ( \frac{A}{E_{1}} \right )+p\left ( E_{2} \right )\cdot p\left ( \frac{A}{E_{2}} \right )}
                   = \frac{\frac{2}{3}\times \frac{1}{2}}{\frac{1}{3}\times \frac{3}{8}+\frac{2}{3}\times \frac{1}{2}}= \frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}= \frac{\frac{1}{3}}{\frac{11}{24}}
                  =\frac{8}{11}  

Posted by

Ravindra Pindel

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