A game consists of tossing a coin $3$ times and nothing its outcome. A boy wins if all tosses give the same outcome and loses otherwise. Find the probability that the boy loses the game.

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Solution

Possible outcomes = $(H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T), (T, T, T)$
Total Possible outcomes $= 8$
Number of outcomes where Hanif loses the game
$= 8 - 2 = 6$
$P (b o y w i l l l o s e t h e g a m e) = \frac{N o . o f o u t c o m e s w h e r e b o y l o s e s}{T o t a l o u t c o m e s}$
$= \frac{6}{8} = \frac{3}{4}$ .

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A game consists of tossing a coin 3 times and nothing its outcome. A boy wins if all tosses give the same outcome and loses otherwise. Find the probability that the boy loses the game.

Possible outcomes = (H,H,H),(H,H,T),(H,T,H),(T,H,H),(T,T,H),(T,H,T),(H,T,T),(T,T,T)Total Possible outcomes =8Number of outcomes where Hanif loses the game =8−2=6 P(boy will lose the game)=No. of outcomes where boy losesTotal outcomes=68=34.

A game consists of tossing a coin $3$ times and nothing its outcome. A boy wins if all tosses give the same outcome and loses otherwise. Find the probability that the boy loses the game.