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.
"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.
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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
(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