14.Probability
easy

Three coins are tossed once. Find the probability of getting exactly $2$ tails.

A

$\frac{3}{8}$

B

$\frac{3}{8}$

C

$\frac{3}{8}$

D

$\frac{3}{8}$

Solution

When three coins are tossed once, the sample space is given by $S =\{ HHH , HHT , HTH , THH , HTT , THT , TTH , TTT \}$

$\therefore$ Accordingly, $n ( S )=8$

It is known that the probability of an event $A$ is given by

$P ( A )=\frac{\text { Number of outcomes favourable to } A }{\text { Total number of possible outcomes }}=\frac{n( A )}{n( S )}$

 Let $H$ be the event of the occurrence of exactly $2$ tails.

Accordingly, $H =\{ HTT ,\,THT, \, TTH \}$

$\therefore P ( H )=\frac{n( H )}{n(S)}=\frac{3}{8}$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.