Three coins are tossed simultaneously. what is the probability of :
$(i)$ getting at most two heads $(i i)$ getting at most two tails.

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Solution

When three coins are tossed simultaneously, $t h e s a m p l e s p a c e b e c o m e s 2^{3} = 8$
$S = (HHH, HHT, HTH, THH, TTH, THT, HTT, TTT)$
Probability of getting at most two heads:
Let $A$ be the event of getting at most two heads.
So, the favourable outcome are $(HHT, HTH, TTT, THH, TTH, THT, HTT)$ .
Therefore, $n (A) = 7$
Hence, $t h e p r o b a b i l i t y o f g e t t i n g a t m o s t t w o h e a d s = \frac{N u m b e r o f f a v o u r a b l e o u t c o m e s}{T o t a l n u m b e r o f o u t c o m e s} = \frac{7}{8} .$
Probability of getting at most two tails:
Let $B$ be the event of getting at most two tails.
So, the favourable outcomes are $(H H H, H H T, H T H, T H H, T T H, T H T, H T T)$
Thus, $n (B) = 7$
Hence, $t h e p r o b a b i l i t y o f g e t t i n g a t m o s t t w o t a i l s = \frac{7}{8}$

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