Answer to Question #302177 in Statistics and Probability for Coudee

Question

1.   Each main bearing cap in an engine contains 4 bolts. The bolts are selected at random without replacement from a parts bin that contains 30 bolts from one supplier and 70 bolts from another.

a.    

b.   What is the probability that a main bearing cap contains all bolts from the same supplier?

 

 

c.   What is the probability that exactly 3 bolts are from the same supplier?

Answer 1

Let $A$ denote the event "bolt from the first supplier". Let $B$ denote the event "bolt from the second supplier".

There are $\dbinom{30+70}{4}=3921225$ possible outcomes.

a. The probability that the number of bolts from each supplier is the same is

$$P(2\ same \ \&\ 2\ other)=P(2A, 2B)$$

$$=\dfrac{\dbinom{30}{2}\dbinom{70}{2}}{\dbinom{100}{4}}=\dfrac{435(2415)}{3921225}=0.267907$$

b.  The probability that a main bearing cap contains all bolts from the same supplier is

$$P(4\ same)=P(4A, 0B)+P(0A, 4B)$$

$$=\dfrac{\dbinom{30}{4}\dbinom{70}{0}}{\dbinom{100}{4}}+\dfrac{\dbinom{30}{0}\dbinom{70}{4}}{\dbinom{100}{4}}$$

$$=\dfrac{ 27405(1)+1( 916895)}{3921225}=0.240818$$

c. the probability that exactly 3 bolts are from the same supplier

$$P(3\ same \ \&\ 1\ other)=P(3A, 1B)+P(1A, 3B)$$

$$=\dfrac{\dbinom{30}{3}\dbinom{70}{1}}{\dbinom{100}{4}}+\dfrac{\dbinom{30}{1}\dbinom{70}{3}}{\dbinom{100}{4}}$$

$$=\dfrac{4060(70)+30(54740)}{3921225}=0.491275$$