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14.Probability
easy
Three coins are tossed once. Find the probability of getting atmost two tails.
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 $J$ be the event of the occurrence of at most $2$ tails.
Accordingly, $J=\{H H H,\, H H T , \,H T H , \,T H H , \,H T T , \,T H T , \, T T H \} ~$
$\therefore P(J)=\frac{n(J)}{n(S)}=\frac{7}{8}$
Standard 11
Mathematics