Question

A coin is tossed twice. If events A and B are defined as :

A = head on first toss, $B =$ head on second toss. Then the probability of $A \cup B =$

• A. $\frac { 1 } { 4 }$
• B. $\frac { 1 } { 2 }$
• C. $\frac { 1 } { 8 }$
• D. $\frac { 3 } { 4 }$

Answer 1

Answer: (D)

$\frac { 3 } { 4 }$

Answer 2

Total number of ways $= ( H H , H T , T H , T T )$

$P$ (Head on first toss) $= \frac { 2 } { 4 } = \frac { 1 } { 2 } = P ( A )$

$P$ (Head on second toss) $= \frac { 2 } { 4 } = \frac { 1 } { 2 } = P ( B )$

$P$ (Head on both toss) $= \frac { 1 } { 4 } = P ( A \cap B )$

Hence required probability is,

$P ( A \cup B ) = P ( A ) + P ( B ) - P ( A \cap B ) = \frac { 1 } { 2 } + \frac { 1 } { 2 } - \frac { 1 } { 4 } = \frac { 3 } { 4 }$ .