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. $\frac{11}{16}$
• B. $\frac{5}{16}$
• C. $\frac{5}{8}$
• D. $\frac{1}{16}$

Answer 1

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

$P(sum>15)=P(5,5,5,5)+P(5,5,5,3)+P(5,5,3,3)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}=\frac{11}{16}$