Test: Binomial Distribution - Commerce MCQ

Mean= E(X) = 
= 0×0.3+1×0.7
=0.7

= 02×0.3+12×0.7 =0.7
Now, ∴ Var(X)= E(X2)-(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21

In a single toss, P(success) = 2/6 = 1/3 and P(non-success) = (1 − 1/3) = 2/3.
P(X=0) = P(non-success in the 1st draw and non-success in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and non-success in the 2nd) or (non-success in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.

Probability of success(p) = ⅗
q = 1 - p
= 1 - (⅗)  ⇒ ⅖
Probability of 2 hit in 5 attempts = 5C2 (⅗)² (⅖)³ 
= 144/625

Mean = np = 3, S.D. (npq)^½ = 3/2

p = ¼            n = 12
Hence, binomial distribution is (p+q)ⁿ = (¾ + ¼)¹²
 

Probability of getting first non-defective bulb =90/100
Probablity of getting second non-defective bulb =89/99
Probablity of getting third non-defective bulb =88/98
Probablity of getting fourth non-defective bulb =87/97
Probablity of getting fifth non-defective bulb =86/96
So, probablity of getting all non-defective bulbs in a sample of 5 bulbs =
(90/100)∗(89/99)∗(88/98)∗(87/97)∗(86/96)
= closest option is b

Let X˜B(n,p) be a binomial variate with mean 4 and variance 3. Then,
np=4,and npq=3
⇒q=3/4,p=1/4 and n=16
∴P(X=r)=.16Cr(1/4)^r (3/4)^(16−r) ,r=0,1,2,....,16
⇒P(X=6)= 16C6(1/4)⁶ ((3)/(4))¹⁰

Probability of an egg being defective =10/100=110
So, probability of an egg being non-defective=1−0.1=0.9
10 eggs are drawn successively with replacement.
So, the probability of getting no defective egg =(0.9)¹⁰
Hence, the probability that there is at least one defective egg = 1−(0.9)¹⁰