Answer to Question #118622 in Statistics and Probability for Robin

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 $=0.6144.$