Question
Three different coins are tossed simultaneously. Find the probability of getting exactly one head.
✓
Answer
Number of possible out comes $=8$
$\{( H , H , H ),( H , H , T ),( H , T , H ),( T , H , H ),( H , T$, $T ),( T , H , T )( T , T , H ) \&(T, T , T )\}$
Number of favourable outcomes $=3$
$\{( H , T , T ),( T , H , T ),( T , T , H )\}$
$\therefore$ Probability $( p )=\frac{\text { favourable outcomes }}{\text { possible outcomes }}=\frac{3}{8}$
Hence, probability of getting exactly one head
$
=\frac{3}{8}
$
$\{( H , H , H ),( H , H , T ),( H , T , H ),( T , H , H ),( H , T$, $T ),( T , H , T )( T , T , H ) \&(T, T , T )\}$
Number of favourable outcomes $=3$
$\{( H , T , T ),( T , H , T ),( T , T , H )\}$
$\therefore$ Probability $( p )=\frac{\text { favourable outcomes }}{\text { possible outcomes }}=\frac{3}{8}$
Hence, probability of getting exactly one head
$
=\frac{3}{8}
$
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