Question

A coin whose faces marked 2 and 3 is thrown 5 times, then chance of obtaining a total of 12 is

• A. $\frac{5}{16}$
• B. $\frac{5}{8}$
• C. $\frac{5}{32}$
• D. $\frac{5}{24}$

Answer 1

The correct option is A $\frac{5}{16}$

It is given that the coin has faces 2 and 3, tossed five times.

To get sum 12 we need 2,2,2,3,3 in any combination.

So, in the 5 throws, there need to be 3 2's. Automatically other 2 will be 3's since there is no other possibility.

So, it boils down to choosing 3 from 5.

Therefore P(getting sum 12) $=5C_3{\left(\frac{1}{2}\right)}^5=\frac{5}{16}.$