A probability task with two parts

goosepane

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Hello everyone, i don't understand how to solve this basic tasks, if someone could solve them and explain the solution that would be very nice. (Thanks in advance).
1.Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B are the next events: A: "Four items are non-defective" B: "two items are defective and two items are non-defective Find the probability of events A and B.
(Sorry for bad english)

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Jomo

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Hello everyone, i don't understand how to solve this basic tasks, if someone could solve them and explain the solution that would be very nice.
I am sorry but we do not solve problems for students on this site. You need to show some attempt to the solution and we will guide you to the answer by givings hints. So please show us your work and we will go from there. Thanks.
 

mmm4444bot

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Don't post lists of homework exercises with no efforts shown. Please read the forum's submission guidelines; you can start with this summary. Thank you!

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goosepane

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I am sorry but we do not solve problems for students on this site. You need to show some attempt to the solution and we will guide you to the answer by givings hints. So please show us your work and we will go from there. Thanks.
Okay, i figured out that we can solve 1.(A) by using combinatorics formula, so it will be 16C4/11C4 which equals 330/1820=33/182=0.181 i guess this is the right answer, but how do i solve b?
 

pka

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1.Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B are the next events: A: "Four items are non-defective" B: "two items are defective and two items are non-defective Find the probability of events A and B.
Have you read the posting guidelines? If so where is your attempts at working these. Seeing what you have done helps us help you.
Does 1) mean to find the probability of A then that of B? Or is it joint probability?
 

Dr.Peterson

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1.Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B are the next events: A: "Four items are non-defective" B: "two items are defective and two items are non-defective Find the probability of events A and B.
(Sorry for bad english)
Okay, i figured out that we can solve 1.(A) by using combinatorics formula, so it will be 16C4/11C4 which equals 330/1820=33/182=0.181 i guess this is the right answer, but how do i solve b?
First, here is a better version in English of this problem as I understand it:

Let 4 items be chosen at random from a lot containing 16 items of which 5 are defective. Let A and B be the following events: A= "Four items are non-defective", B= "two items are defective and two items are non-defective. Find the probability of each of the events A and B.​

You are saying that Pr(A) = 16C4/11C4; I think you meant 11C4/16C4 = 330/1820. This is the number of ways to chose 4 of the 11 non-defective items, over the number of ways to choose any 4 items. That is correct.

To find Pr(B) by the same technique, your numerator has to be the number of ways to choose 2 from the 5 defective and 2 from the 11 non-defective. Can you see how to do that? Make an attempt.
 
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