Question

Four unbiased coins are tossed simultaneously. The probability of getting at most 2 tails is

• A.
  $\frac{9}{16}$
• B.
  $\frac{11}{16}$
• C.
  $\frac{7}{16}$
• D.
  $\frac{5}{8}$

Answer 1

The correct option is B

$\frac{11}{16}$
When four coins are tossed simultaneously, the total number of possible outcomes is $2^4=16.$

Let A be the event of getting at most 2 tails i.e., 0 tail, 1 tail or 2 tails.

Now, the favourable outcomes are HHHH, THHH, HTHH, HHTH, HHHT, TTHH, THTH, THHT, HTTH, HTHT, HHTT.

$\therefore$Number of favourable outcomes = 11

$\therefore P\left(A\right)=\frac{\text{Number of outcomes favourable to A}}{\text{Total number of outcomes}}=\frac{11}{16}$


Hence, the correct answer is option b.