BITSAT 2018 Slot 2 Solved Paper

Mean (m) for BD=np
and variance (σ2) for BD=npq
Given, np+npq=2.7 and n=3
∴np(1+q)=2.7⇒p(1+q)=

2.7

3

⇒p(1+q)=0.9
⇒(1−q)(1+q)=0.9
[∵ for BD probability standard deviation (SD), p+q=1 ]
⇒1−q2=0.9
⇒q2=0.1⇒q=±0.3
∴q=0.3
[∵q=−0.3 can't possible as 0≤q≤1]
⇒p=0.7 and q=0.3
∴B.D. =(0.3+0.7)5
Hence, option (b) is correct.