A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting
(i) exactly 5 successes
(ii) at least 5 successes
(iii) at most 5 successes
(i) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
As the die is thrown 6 times the total number of outcomes will be 
.
And we know that the favourable outcomes of getting exactly 5 successes will be, either getting 2, 4 or 6 i.e., 1/6 probability of each, total, 
probability, p = 
,q = 
The probability of success is 
and of failure is also 
.
Thus, the probability of getting exactly 5 successes will be
⇒ 6C5
⇒ 6C5.
⇒ 
(ii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
As the die is thrown 6 times the total number of outcomes will be 
.
And we know that the favourable outcomes of getting at least 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, 
probability, p = 
, q = 
The probability of success is 
and of failure is also 
.
Thus, the probability of getting at least 5 successes will be
⇒ (6C5 + 6C6) 
⇒ (6C5 + 6C6).
⇒ 
(iii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
As the die is thrown 6 times the total number of outcomes will be 
.
And we know that the favourable outcomes of getting at most 5 successes will be, either getting 2, 4 or 6 i.e, 1/6 probability of each, total, 
probability of success .
The probability of success is 
and of failure is also 
.
Thus, the probability of getting at most 5 successes will be
⇒ (6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5). 
⇒ (6C0 + 6C1 + 6C2 + 6C3 + 6C4 + 6C5).
⇒ 