Question

Four unbiased coins are tossed simultaneously. The probability that exactly two heads occur will be

• A.
  $\frac{1}{16}$
• B.
  $\frac{3}{8}$
• C.
  $\frac{3}{16}$
• D.
  $\frac{5}{16}$

Answer 1

The correct option is B

$\frac{3}{8}$
When four coins are tossed simultaneously, the total number of possible outcomes is$2^4=16.$

The favourable outcomes having exactly two heads are HHTT, HTHT, HTTH, THHT, TTHH, THTH.

$\therefore$ Number of favourable outcomes = 6
$\therefore P\left(\text{Exactly 2 heads}\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{6}{16}=\frac{3}{8}$

Hence, the correct answer is option b.