Probability of getting an odd number in one trial =

3

6

=

1

2

=p (say)
Probability of getting an even number in one trial =

3

6

=

1

2

=q (say)
X has a binomial distribution with n=6,p=

1

2

,q=

1

2

P(X=r)=nCrqn−rpr,
(i) P (5 successes) = P(X = 5) =6C5qp5=6×

1

2

×

1

25

=

3

32

(ii) P(at least 5 successes) = P(X = 5) + P(X = 6) =

3

32

+6C6q0p6 [Using Part (i)]
=

3

32

+1×1×

1

26

=

7

64

(iii) P (at most 5 successes) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
=1−P(X=6) [∵∑P(X)=1]
=1−

1

64

=

63

64

[Using part (ii)]