Question

three balls are to be randomly selected without replacement from and you are and consisting 20 balls numbered 1 to 20 every bad that at least one of the Dragon Ball has a number as large as or larger than 17 what is the probability of event win the bet​

Answer 1

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

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 = 4C1∗16C2+4C2∗16C1+4C320C3=0.5084C1∗16C2+4C2∗16C1+4C320C3=0.508