Solution to 1. Each main bearing cap in an engine contains 4 bolts. The bolts are selected … - Sikademy
Author Image

Archangel Macsika

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?

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-4-stid-46-sqid-1819-qpid-289