Question

Q1. Obtain the probability distribution of the number of aces in simultaneous throw of two unbiased dice.

Q2. Machine produces an average of 20% defective bolts. A batch is accepted if a sample of 5 bolts taken from that batch contains no defective and rejected it the sample contains 3 or more defectives. In other cases, a second sample is taken. What is the probability that the second sample is required ?

Answer 1

Q1.Two cards are drawn successively without replacement from a wellshuffled deck of 52 cards. Find the probability distribution of the number of aces.

Let X denote the number of aces in a sample of 2 cards drawn from a wellshuffled deck of 52 cards. Then X can take the values 0, 1 and 2.

$$P(X=0)=P(no \ ace)={48 \over 52}\cdot{47 \over 51}={188 \over 221}$$

$$P(X=1)=P(one \ ace)={4 \over 52}\cdot{48 \over 51}+{48 \over 52}\cdot{4 \over 51}={32 \over 221}$$

$$P(X=2)=P(two \ aces)={4 \over 52}\cdot{3 \over 51}={1 \over 221}$$

The probability distribution of X is given by

$$\begin{matrix} X=x & 0 & 1 & 2 \\ \\ p(X=x) & \dfrac{188}{221} & \dfrac{32}{221} & \dfrac{1}{221} \end{matrix}$$

Q2. Let X denote the number of defective bolts in a sample of n bolts: $X\sim Bin(n,p)$

$$P(X=x)=\binom{n}{k}p^x(1-p)^{n-x}$$

Given $p=0.2,n=5.$

$$P(X=1)+P(X=2)=$$

$$=\binom{5}{1}(0.2)^1(1-0.2)^{5-1}+\binom{5}{2}(0.2)^2(1-0.2)^{5-2}=$$

$$=5(0.2)(0.8)^4+10(0.2)^2(0.8)^3=0.6144$$

Probability that a second sample is required $=0.6144.$