Discreate Random Variable (Rose_8th Edition)

Question

Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number as large as or larger than 17, what is the probability that we win the bet?

My Approach through random variables:

let X is the random variable defined as:

X = number of balls having the number on them >= 17 in three drawn.

In the question, they have asked for at least 1 then we can place the values of X are 1,2,3;

P(X = 1 ) = (4/20) * (16/19) * (15 / 18)  =  0.14035

P(X = 2 ) = (4/20) * (3/19) * (16 / 18) = 0.02807

P(X = 3 ) = (4/20) * (3/19) * (2 / 18) = 0.00350

Prob of wining bet = P(X=1) + P(X=2) + P(X=3)

= 0.17192

but the answer is 0.508

what is wrong with this approach.

Comments

https://math.stackexchange.com/questions/2767202/probability-that-if-three-balls-numbered-from-1-20-are-selected-without-replacem

Answer 1

For P(X=1), I think by trying to do 4C1/20C1 = 4/20 you are selecting 4 balls out of 20, which can lead to the selection of any 4 balls which may be numbered between 1-20.

The correct approach would be as follows: For P(x=1), (Select one from 4 balls [1-4] and then select 2 from the remaining [1-16].) /  (Total ways to select 3 balls out of 20.)

Similarly for P(x=2) and P(x=3),

 

Hence, Probability = $\frac{4C1*16C2 +4C2*16C1 + 4C3}{20C3} = 0.508$

Comments

Thanks for making it simple.!!!

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for without replacement combination is used

Answer 2

Let X be the maximum number that appears in any of the three balls.

Hence,

 

P(X>=17)=P(X=17)+P(X=18)+P(X=19)+P(X=20)=1-P(X<=16)=1-(16C3/20C3)=0.508