A die is thrown 4 times. ‘Getting a 1 or a 6’ is considered a success, Find the probability of getting
(i) exactly 3 successes
(ii) at least 2 successes
(iii) at most 2 successes
Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
We know that the favourable outcomes of getting exactly 3 successes will be, either getting 1 or a 6 i.e, total, 
probability
The probability of success is 
and of failure is 
.
Thus, the probability of getting exactly 3 successes will be
⇒ (4C3) 
⇒ (4C3).
⇒ 
(ii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
We know that the favourable outcomes of getting at least 2 successes will be, either getting 1 or a 6 i.e, total, 
probability
The probability of success is 
and of failure is 
.
Thus, the probability of getting at least 2 successes will be
⇒ (4C2) ) 
+ (4C3) ) 
+ (4C4) 
⇒ 
⇒ 
(iii) Using Bernoulli’s Trial P(Success=x) = nCx.px.q(n-x)
x=0, 1, 2, ………n and q = (1-p)
We know that the favourable outcomes of getting at most 2 successes will be, either getting 1 or a 6 i.e, total, 
probability
The probability of success is 
and of failure is 
.
Thus, the probability of getting at most 2 successes will be
⇒ 
⇒ 