7 white balls and 3 black balls are placed in a row at random. The pro

The black ball will not be adjacent if there is a white ball in the middle.

First, place all the 7 white balls, and there will be 8 spaces available for 3 black balls.

Desired result: \(^{8}\mathrm{C_3}\)

Total result: \(^{10}\mathrm{C_3}\)

Probability: \(\frac{Desired}{ Total}\)

=> {8*7*6} / {10*9*8}

=> \(\frac{7}{15}\)

Answer B

Solution:

The number of ways 7 white balls and 3 black balls can be placed in a row is 10! / (7!3!) = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120.

Now, let’s examine the number of ways we can place the black balls in the row so that none of them are adjacent.

Case 1: If all the 3 black balls are somewhere between the white balls (i.e., none of black balls is first or last ball in the row of the 10 balls), then the number of ways is the same as the number of ways of writing the number 7 (the number of white balls) as a sum of 4 positive integers (for example, the arrangement WBWWWBWBWW is same as expressing 7 as 1 + 3 + 1 + 2). Let’s list them (note: we will first list 7 as a sum of 4 addends in ascending order and the total number of ways we can permute the 4 addends):

7 = 1 + 1 + 1 + 4 → Total number of ways = 4!/3! = 4

7 = 1 + 1 + 2 + 3 → Total number of ways = 4!/2! = 12

7 = 1 + 2 + 2 + 2 → Total number of ways = 4!/3! = 4

Therefore, the total number of ways 7 can be expressed as a sum of 4 positive integers is 4 + 12 + 4 = 20.

Case 2: If one of the 3 black balls is the first ball but neither of the other 2 black balls is the last ball of the row, then the number of ways is the same as the number of ways of writing 7 as a sum of 3 positive integers (for example, the arrangement BWWWWBWBWW is same as expressing 7 as 4 + 1 + 2). Let’s list them (using the same strategy as in case 1):

7 = 1 + 1 + 5 → Total number of ways = 3!/2! = 3

7 = 1 + 2 + 4 → Total number of ways = 3! = 6

7 = 1 + 3 + 3 → Total number of ways = 3!/2! = 3

7 = 2 + 2 + 3 → Total number of ways = 3!/2! = 3

Therefore, the total number of ways 7 can be expressed as a sum of 3 positive integers is 3 + 6 + 3 + 3 = 15.

Case 3: If one of the 3 blacks is the last ball but neither of the other 2 black balls is the first ball of the row, then the number of ways is the same as in case 2. That is, there are 15 ways this can happen.

Case 4: If a black ball is the first ball and another ball ball is the last ball of the row, then the number of ways is the same as the number of ways of writing 7 as a sum of 2 positive integers (for example, the arrangement BWWWWBWWWB is same as expressing 7 as 4 + 3). Let’s list them (using the same strategy as in case 1):

7 = 1 + 6 → Total number of ways = 2! = 2

7 = 2 + 5 → Total number of ways = 2! = 2

7 = 3 + 4 → Total number of ways = 2! = 2

Therefore, the total number of ways 7 can be expressed as a sum of 3 positive integers is 2 + 2 + 2 = 6.

Adding up the number of ways in the 4 cases above, the total number of ways the black balls in the row of 10 black and white balls such that none of the black balls are adjacent is 20 + 15 + 15 + 6 = 56. Therefore, the probability is 56/120 = 7/15.

Alternate Solution:

The number of ways 7 white balls and 3 black balls can be placed in a row is 10! / (7!3!) = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120.

Now, let’s examine the number of ways we can place the black balls in the row so that at least two of them are adjacent.

If all three black balls are adjacent, then we are arranging the objects [BBB]-W-W-W-W-W-W-W; i.e. 8 objects, 7 of which are indistinguishable. There are 8!/7! = 8 ways to arrange these objects.

If at least two black balls are adjacent, then we are arranging the objects [BB]-B-W-W-W-W-W-W-W; i.e. 9 objects, 7 of which are indistinguishable (notice that [BB] is considered as a single object, thus it can be distinguished from B). There are 9!/7! = 9 x 8 = 72 ways to arrange these objects.

Notice that the 72 ways of arrangements above actually include the ways where all three black balls are adjacent, since an arrangement can be [BB]-B-W-W-W-W-W-W-W. Further, each arrangement where all three black balls are adjacent is counted twice, as the above arrangement is counted again as B-[BB]-W-W-W-W-W-W-W. Thus, the number of arrangements where two black balls are adjacent but three black balls are not adjacent is 72 - 2 x 8 = 56.

Adding the arrangements where at most two and all three black balls are adjacent, we see that in 56 + 8 = 64 of the arrangements, at least two black balls are adjacent. Then, in 120 - 64 = 56 of the arrangements, no two black balls are adjacent. Since there are 120 total ways to arrange the balls and 56 of them are favorable, the probability that no two black balls are adjacent is 56/120 = 7/15.

Answer: B

Alternatively, instead of placing white balls first, we can place the black balls first:

_b_b_b_

Place a white ball between the B's. This leaves 5 white balls remaining to be placed anywhere in the 4 spots.

Using the sticks method, the black balls represent the sticks.

So 5 balls + 3 "sticks" = 8

8!/5!3!= 56

Ways to arrange all 10 balls without restrictions

10!/7!3! to take into account identical balls in the permutation =

120

Probability 56/120 = 7/15

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