Question

In a single throw of two dice find the probability that neither a doublet nor a total of 8 will appear.

• A. $\frac{1}{21}$
• B. $\frac{1}{121}$
• C. $\frac{1}{18}$
• D. $\frac{13}{18}$

Answer 1

The correct option is D $\frac{13}{18}$

Let 'A' be the event that a double occur and b be the

event that total on 2 dice is 9.

Let 's' be the total no. of outcomes, i.e. n(s) = 36

$\therefore A=\left\{\right(1,1),(2,2),(3,3),(4,4),(5,5),(6,6\left)\right\}$

$B=\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}$

$\Rightarrow n\left(A\right)=6$ & $n\left(B\right)=5$ & $n(A\cap B)=1$

$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{36}$ & $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{5}{36}$ & $P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{1}{36}$

$\therefore$ Probability that neither A nor B occur $=1-P(A\cup B)$

$=1-\left\{P\right(A)+P(B)-P(A\cap B\left)\right\}$

$=1-\{\frac{6}{36}+\frac{5}{36}-\frac{1}{36}\}$

$=\frac{26}{36}$ $=\frac{13}{18}$