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14.Probability
easy
Three coins are tossed once. Find the probability of getting atleast $2$ heads.
A
$\frac{1}{2}$
B
$\frac{1}{2}$
C
$\frac{1}{2}$
D
$\frac{1}{2}$
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 $D$ be the event of the occurrence of at least $2$ heads.
Accordingly, $D =\{ HHH ,\, HHT \,, HTH \,, THH \}$
$\therefore P(D)=\frac{n(D)}{n(S)}=\frac{4}{8}=\frac{1}{2}$
Standard 11
Mathematics
