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1. In a certain lottery, three white balls are drawn (at random) from ten balls numbered from 1 to 10, and one red SuperBall is drawn (at random) from ten balls numbered from 11 to 20. When you buy a ticket, you select three numbers from 1 to 10 and one number from 11 to 20. To win the jackpot, the numbers on your ticket must match the three white balls and the red SuperBall. (You don't need to match the white balls in order).
If you buy a ticket, what is your probability of winning the jackpot?

 

2. In a certain lottery, three white balls are drawn (at random) from ten balls numbered from 1 to 10, and one red SuperBall is drawn (at random) from ten balls numbered from 11 to 20. When you buy a ticket, you select three numbers from 1 to 10 and one number from 11 to 20. To win a super prize, the numbers on your ticket must match at least two of the white balls or must match the red SuperBall.
If you buy a ticket, what is your probability of winning a super prize?

 Feb 29, 2020
 #1
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1. The probability is 1/C(20,3) = 1/1140.

 

2  nCr(20, 2) = 190 ways of selecting 2 from 20

nCr(10, 1)    =   10 ways of selecting 1 from 10

 

To find the over all probability of winning,

Calculate the Harmonic Mean (19) and divide by 2

Equivalent to

1/ (1//190 + 1/10) =9.5

 

So the probability is 1/9.5 = 2/19.

 Feb 29, 2020
 #2
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Hello, for 1., the answer was 1/1200, and for 2., the answer was not 2/19. I don't understand where you did it wrong, and if you would like to re-try or if anyone else can attempt it, that would be great. Thank you for trying!

somehelpplease  Mar 1, 2020
 #3
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1.  The probability your first ticket matches one of the white balls is 3/10.  Then the second ticket has a probability of matching of 2/9 (since there are only 9 white balls left). Then the probability of the third ticket matching is 1/8.  The probability of matching the red ball is 1/10.  Hence, the overall probability is (3/10)*(2/9)*(1/8)*(1/10) = 1/1200.

 

2.  I think this is as follows:

 

p(exactly 3 white) = (3/10)*(2/9)*(1/8) =  1/120

p(exactly 2 white) = (3/10)*(2/9)*(7/8) = 7/120

p(1 red) = 1/10

So probability(at least 2 white or 1 red) =  1/120 + 7/120 + 1/10= 1/6

 Mar 1, 2020
edited by Alan  Mar 1, 2020
edited by Alan  Mar 1, 2020
 #4
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I'm sorry Alan, that wasn't the correct answer. I don't think you read the question right, but thank you for trying!

somehelpplease  Mar 1, 2020
 #5
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Well, it's quite possible I read the question incorrectly (I assume you refer to part 2). It's also possible my logic is wrong. However, it would be nice to know in what way I read it incorrectly. I certainly misread it at first as I initially assumed part 2 required at least two white balls and the red ball. I then noticed the question specified at least two white balls or the red ball. Perhaps you can explain which bit I read incorrectly (or else supply the correct answer so I might be able to work out the error myself).

Alan  Mar 1, 2020
 #6
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It says 2 white balls, or a red ball. Not both.

Guest Mar 1, 2020
 #7
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The word "or" is ambiguous. 

It can mean inclusive or (one or the other or both), which is what I took it to mean.

It can mean exclusive or (one or the other but not both).

 

If exclusive or was meant here then the probability becomes:

 

p(at least 2 white)*(1-p(red)) + p(red)*(1-p(at least two white))   so:   (1/15)*(1-1/10) + (1/10)*(1-1/15) = 23/150

 

(The above numerical value assumes I've calculated the individual probabilities correctly!).

Alan  Mar 2, 2020
 #8
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Sorry, Alan, I should have specified. It is inclusive, and I'm sorry for the confusion. As for the right answer, I don't know it, but when I get it I will make sure to send you the explanation for the answer. Thank you for trying to help!

somehelpplease  Mar 2, 2020

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