Question

If one die is thrown twice, then what will be the probability of getting 9 as sum of the two outcomes? (Given that the outcome on the die is 5 in the first throw.)

• A. $\frac{1}{6}$
• B. $\frac{1}{4}$
• C. $\frac{1}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{6}$

$\mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{the}\mathrm{following}\mathrm{events}:A:‘\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{outcomes}\mathrm{is}9’.B:‘\mathrm{die}\mathrm{shows}5\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{throw}’.\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}P\left(A|B\right).\mathrm{There}\mathrm{are}36\mathrm{elements}\mathrm{in}\mathrm{the}\mathrm{sample}\mathrm{space}S=\left(\left(x,y\right):x,y=1,2,3,4,5,6\right\}.A=\left(\left(3,6\right),\left(4,5\right),\left(5,4\right),\left(6,3\right)\right\}B=\left(\left(5,y\right):y=1,2,3,4,5,6\right\}\mathrm{Thus},\mathrm{there}\mathrm{are}6\mathrm{elements}\mathrm{in}B.A\cap B=\left(\right(5,4\left)\right\}\mathrm{Now},P\left(B\right)=\frac{6}{36}\mathrm{and}P\left(A\cap B\right)=\frac{1}{36}\mathrm{Then},P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\Rightarrow P\left(A|B\right)=\frac{\frac{1}{36}}{\frac{6}{36}}=\frac{1}{6}$