Answer to Question #270426 in Statistics and Probability for Light

Question

The quality control manager of WHEAT Bakery has observed (over a long period of time) that 2% of loaves of bread baked in the evening are either underweight or over weight and they must not be sold. If she took a sample of 6 such loaves of bread for inspection, find the probability that:

Exactly 2 are unacceptable.

At most 3 are unacceptable.

Between 3 and 5 (inclusive) are no unacceptable.

(b) How many loaves of bread would be expected to be unacceptable from a sample of 100 such loaves?

(c) Suppose that A and B are events in a sample space S such that p ( A )  0 .4 , p(A  B)  0.7 and p(B)  k

(i) What is the numerical value of k for which events A and B are independent.

(ii) What is the numerical value of k for which A and B are mutually exclusive.

Answer 1

(a) Let $X=$ the number of unacceptable loaves of bread: $X\sim Bin(n, p).$

Given $n=6, p=0.02,q=1-0.02=0.98$

(i) Exactly 2 are unacceptable.

$$P(X=2)=\dbinom{6}{2}(0.02)^2(0.98)^{6-2}$$

$$=0.00553420896$$

(ii) At most 3 are unacceptable.

$$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)$$

$$+P(X=3)=\dbinom{6}{0}(0.02)^0(0.98)^{6-0}$$

$$+\dbinom{6}{1}(0.02)^1(0.98)^{6-1}+\dbinom{6}{2}(0.02)^2(0.98)^{6-2}$$

$$+\dbinom{6}{3}(0.02)^3(0.98)^{6-3}=0.885842380864$$

$$+0.108470495616+0.00553420896$$

$$+0.00015059072=0.99999767616$$

(iii) Between 3 and 5 (inclusive) are no unacceptable. That is at most 2 are unacceptable or Exactly 6 are unacceptable.

$$P(X\leq 2\ or\ X=6)=P(X=0)+P(X=1)$$

$$+P(X=2)+P(X=6)$$

$$=\dbinom{6}{0}(0.02)^0(0.98)^{6-0}+\dbinom{6}{1}(0.02)^1(0.98)^{6-1}$$

$$+\dbinom{6}{2}(0.02)^2(0.98)^{6-2}+\dbinom{6}{6}(0.02)^6(0.98)^{6-6}$$

$$=0.885842380864+0.108470495616$$

$$+0.00553420896+0.000000000064$$

$$=0.999847085504$$

(b)

Given $n=100, p=0.02$

$$E(X)=np=100(0.02)=2$$

2 loaves of bread would be expected to be unacceptable from a sample of 100 such loaves.

(c)

Given $P(A)=0.4, P(A\cup B)=0.7, P(B)=k$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$P(A\cap B)=P(A)+P(B)-P(A\cup B)$$

$$P(A\cap B)=0.4+k-0.7=k-0.3$$

(i)

$$P(A\cap B)=P(A)P(B)$$

$$k-0.3=0.4k$$

$$0.6k=0.3$$

$$k=0.5$$

(ii)

$$P(A\cup B)=P(A)+P(B)$$

$$0.7=0.3+k$$

$$k=0.4$$