A coin is tossed 10 times. Find the probability of getting
(i) 2 Heads
(ii) Even numbers of heads

Explanation

(i) Total no. of outcomes when coin is tossed 10 times is `2^(10)`=1024
When 2 heads are there, this mean 8 tails also there.
i.e. we want the no. of ways in which
HHTTTTTTTT can be arranged is `(10!)/(2!8!)=45`
required probability `=45/1024`
(ii) (0 heads and 10 tails) or (2 heads and 8 tails) or (4 heads and 6 tails) or ( 6 heads and 4 tails) or (8 heads and 2 tails) or 10 heads and 10 tails0
`=((10!)/(0!10!)+(10!)/(2!8!)+(10!)/(4!6!)+(10!)/(8!2!)+(10!)/(10!0!))/(2^10)`
`(""^10C_0+""^10C_2+""^10C_4+""^10C_6+""^10C_8+""^10C_10)/(2^10)=(2^(10-1))/(2^10)`
`2^9/2^10=1/2`