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.

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Rita Miller · Accepted Answer

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 · Accepted Answer

Probability to choose 3 not previously used balls from 15 when 9 are not previously used:  9⋅8⋅715⋅14⋅13  Conditionally on this, probability that these 3 balls were not played with because they would have been chosen in the first phase:  12⋅11⋅1015⋅14⋅13  Thus, the desired probability is  12⋅11⋅10⋅9⋅8⋅7(15⋅14⋅13)2≈0.893.

Vasquez · Accepted Answer

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 &amp;amp; all 3 used. So P(E1)=∑n=03(9n)⋅(63−n)(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(9−n3)(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)=(9−1c315c3) [9-1 because out of 9 unused balls, 1 ball is chosen] Now total probability =P(E1)⋅P(E2) =∑n=03(9n)⋅(63−n)(153)⋅(9−n3)(153)=0.089

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

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