Question

A coin whose faces are marked $3,5$ is tossed $4$ times. The probability that the sum of the numbers thrown is greater than $15$

• A. $\frac{13}{16}$
• B. $\frac{5}{16}$
• C. $\frac{5}{8}$
• D. $\frac{11}{16}$

Answer 1

The correct option is D $\frac{11}{16}$

Total number of possible outcomes in $4$ tosses $=2^4=16$
$P(\text{sum}>15)=P(5,5,5,5)+P(5,5,5,3)+P(5,5,3,3)$
$=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}=\frac{11}{16}$