There are 15 tennis balls in a box, of which 9 have not previously been...

osnomu3

osnomu3

Answered

2021-12-31

There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

Answer & Explanation

Rita Miller

Rita Miller

Expert

2022-01-01Added 28 answers

Step 1
We consider the four possible cases of the initial draw.
Case 0: No used balls are drawn.
p0=(93)(153)
Case 1: 1 used ball is drawn.
p1=(92)6(153)
Case 2: 2 used balls are drawn.
p2=9(62)(153)
Case 3: 3 used balls are drawn.
p3=(63)(153)
Step 2
We evaluate the probabilities and outcomes of each case.
Case 0: p0=.1846, 6 new balls, 9 used left over.
Case 1: p1=.4747, 7 new balls, 8 used.
Case 2: p2=.2967, 8 new balls, 7 used.
Case 3: p3=.044, 9 new balls, 6 used.
Step 3
We assess the probability of each case times the probability that no used balls turn up in the second draw, and add everything up.
p0(63)(153)+p1(73)(153)+p2(83)(153)+p3(93)(153)
Step 4
.1846.044+.4747.0769+.2967.1231+.044.1846=.0893
Timothy Wolff

Timothy Wolff

Expert

2022-01-02Added 26 answers

Probability to choose 3 not previously used balls from 15 when 9 are not previously used:
987151413
Conditionally on this, probability that these 3 balls were not played with because they would have been chosen in the first phase:
121110151413
Thus, the desired probability is
121110987(151413)20.893.
Vasquez

Vasquez

Expert

2022-01-07Added 457 answers

We have 15 balls out of which 9 balls have not been previously used, so the remaining 6 balls are already used.
Let E1 be the first event when the 3 balls are randomly chosen and played with.
Now when choosing 3 balls, the combination can be choosing all 3 unused, 2 unused and 1 used, 1 unused and 2 used & all 3 used.
So P(E1)=n=03(9n)(63n)(153)
Here n is the no of unused balls. For eg:- when n is 1, we can choose 1 unused ball from 9 unused in (9c1) ways and remaining 2 from 6 used balls in (6c2) ways .
Now let E2 be event of picking 3 unused balls after E1.
Then P(E2)=n=03(9n3)(153). Here n is the no of unused balls chosen after event E1. If suppose in E1, 1 unused and 2 used balls were chosen then P(E2)=(91c315c3) [9-1 because out of 9 unused balls, 1 ball is chosen]
Now total probability =P(E1)P(E2)
=n=03(9n)(63n)(153)(9n3)(153)=0.089

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