Question

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Q.7. A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Answer 1

Dear Student,

$$\begin{gathered} LetSbetheeventthatthesumofthenumbersappearingin2throwsofthediceis6.Then,S=\left\{\right(1,5),(2,4),(3,3),(4,2),(5,1\left)\right\}=5outcomes. \\ LetFbetheeventthat4appearsatleastonceonthedice.Then,F=\left\{\right(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4),(5,4),(4,5),(6,4),(4,6\left)\right\}=11outcomes. \\ Theoutcomespossibleare\left\{\right(1,1),(1,2),...(1,6),(2,1),(2,2),...(2,6),(3,1),(3,2),...(3,6),(4,1),(4,2),...(4,6),(5,1),(5,2),...(5,6),(6,1),(6,2),...(6,6\left)\right\}.Thus,thereare36possibleoutcomes. \\ Now,S\cap F=\left\{\right(2,4),(4,2\left)\right\} \\ \therefore Conditionalprobabilitythat4hasappearedatleastoncegiventhatthesumofthenumbersintwothrowsofthediceis6=P(F/S)=\frac{P(S\cap F)}{P\left(S\right)}=\frac{{\displaystyle \frac{2}{36}}}{{\displaystyle \frac{5}{36}}}=\frac{2}{5}=0.4 \end{gathered}$$