Three coins are tossed once. probability of getting at most 2 heads.

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14.Probability

easy

Three coins are tossed once. Find the probability of getting at most $2$ heads.

A

$\frac{7}{8}$

B

$\frac{7}{8}$

C

$\frac{7}{8}$

D

$\frac{7}{8}$

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 $E$ be the event of the occurrence of at most $2$ heads.

Accordingly, $E =\{ HHT , \,HTH , \,THH , \,HTT , \,THT \,, TTH , \,TTT \}$

$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{7}{8}$

Standard 11

Mathematics

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