Math: Help with probability
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Math: Help with probability

[From: ] [author: ] [Date: 12-05-22] [Hit: ]
How many ways can a hospital purchase 4 of these sets and receive at least 2 of the defective sets?In a 10-question test with four choices for each question, a student guesses on each item. Find the probability of the student getting six questions correct. Find the probability of the student getting at least eight questions correct.I dont need to know the answer,......
Please help me with these math problems on probability. It's not that i'm lazy to do my own homework, i actually need help in understanding how to do them.

"A shipment of 10 TV sets contains 3 defective sets. How many ways can a hospital purchase 4 of these sets and receive at least 2 of the defective sets?"


and

"In a 10-question test with four choices for each question, a student guesses on each item. Find the probability of the student getting six questions correct. Find the probability of the student getting at least eight questions correct."

I don't need to know the answer, I just really need to know HOW to do it and where to start at.

All help is appreciated. Thanks.

-
banana -

(1) A shipment of 10 TV sets contains 3 defective sets. How many ways can a hospital purchase 4 of these sets and receive at least 2 of the defective sets?

There are 3C2 = 3 ways to receive 2 defective and there are 7C2 = 21 ways to receive 2 non-defective. In, total there are 3 x 21 = 63 ways to receive exactly 2 defective out of 4.

There are 3C3 = 1 way to receive 3 defective and there are 7C1 = 7 ways to receive 1 non-defective. In, total there are 1 x 7 = 7 ways to receive exactly 3 defective out of 4.

These two events are independent, so simply add the results: Answer: 63 + 7 = 70 ways.


(2) In a 10-question test with four choices for each question, a student guesses on each item. Find the probability of the student getting six questions correct.

This is simply Binomial with n=10 and p=0.25

P(6) = 10C6 (0.25^6)(0.75^4) = 0.016222


Find the probability of the student getting at least eight questions correct.

Same as above except sum these 3 Binomial terms:

P(X >=8) = 10C8 (0.25^8)(0.75^2) + 10C9 (0.25^9)(0.75^1) +10C10 (0.25^10)(0.75^0) = 0.0004158

Hope that helps
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