14.Probability
easy

Three coins are tossed once. Find the probability of getting $3 $ heads

A

$\frac{1}{8}$

B

$\frac{1}{8}$

C

$\frac{1}{8}$

D

$\frac{1}{8}$

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 $B$ be the event of the occurrence of $3$ heads. Accordingly, $B=\{H H H\}$

$\therefore $ $P(B)=\frac{n(B)}{n(S)}=\frac{1}{8}$

Standard 11
Mathematics

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