There are 5 black balls and 15 white balls. Balls of the same color
are identical. we take out Randomly all the balls from the box without
return and arrange them in a row.1- Calculate the probability that all the black balls are in adjacent.
2- Calculate the probability that between every two black balls there
will be at least 2 white balls3- Calculate the probability that the third ball in
arrangement is a black ball.
1- There are 16 ways of arranging black balls to be in 1 row to count them as 1 black ball such that C(16,1). Then I think we divide by all the ways of arranging the balls such that 16/20! is probability that all the black balls are in adjacent
2- We take the of competence of sample where there no between black balls any white ball, or in other words same as previous question, all the black balls are in adjacent.
we get (1-16/20!) is the probability that between every two black balls there will be at least 2 white balls
3- Calculate the probability that the third ball in arrangement is a black ball.
We assume the third spot is already a black ball, the number of ways to arrange the other 19 balls is 19!, with the sample space 20! we get 19!/20!= 1/20
Are my answers correct?
Best Answer
Since balls of the same color are indistinguishable, two arrangements are distinguished by which five of the $20$ positions in the row are occupied by black balls. Hence, there are $\binom{20}{5}$ possible arrangements in the sample space, as Thomas Andrews stated in the comments.
If all five black balls are consecutive, the block of black balls must begin in one of the first $16$ positions.
Hence, the probability that all the black balls appear consecutively is $$\frac{16}{\dbinom{20}{5}}$$
The complementary event to the event that there are at least two white balls between each pair of successive black balls is that there is at most one white ball between each pair of successive black balls.
Place the five black balls in a row. This creates six spaces in which to place the white balls, four between successive black balls and two at the ends of the row. $$\square b \square b \square b \square b \square b \square$$ Let $x_i$ be the number of white balls placed in the $i$th such space. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 15 \tag{1}$$ is an equation in the nonnegative integers subject to the constraints that $x_2, x_3, x_4, x_5 \ge 2$ since there must be at least two white balls between any pair of successive black balls.
Let $x_2' = x_2 - 2$, $x_3' = x_3 - 2$, $x_4' = x_4' - 2$, and $x_5' = x_5 - 2$. Then $x_2', x_3', x_4', x_5'$ are nonnegative integers. Substituting $x_2' + 2$ for $x_2$, $x_3' + 2$ for $x_3$, $x_4' + 2$ for $x_4$, and $x_5' + 2$ for $x_5$ in equation $1$ and simplifying yields $$x_1 + x_2' + x_3' + x_4' + x_5' + x_6 = 7 \tag{2}$$ Equation 2 is an equation in the nonnegative integers. A particular solution of equation $2$ corresponds to the placement of $6 - 1 = 5$ addition signs in a row of $7$ ones. For instance, $$1 1 + + 1 + 1 1 1 + 1 +$$ corresponds to the solution $x_1 = 2, x_2 = 0, x_3 = 1, x_4 = 3, x_5 = 1, x_6 = 0$. The number of solutions of equation $2$ in the nonnegative integers is the number of ways five addition signs can be placed in a row of seven ones, which is $$\binom{7 + 6 - 1}{6 - 1} = \binom{12}{5}$$ since we must choose which five of the $12$ positions required for seven ones and five addition signs will be filled with addition signs.
Hence, the probability that there are at least two white balls between each pair of black balls is $$\frac{\dbinom{12}{5}}{\dbinom{20}{5}}$$
In your attempt, you did not account for the fact that balls of the same color are indistinguishable.
Place a black ball in the third position. That leaves four black balls and $15$ white balls to be distributed to the remaining $19$ positions, which can be done in $\binom{19}{4}$ ways. Hence, the probability that there is a black ball in the third position is $$\frac{\dbinom{19}{4}}{\dbinom{20}{5}} = \frac{\dfrac{19!}{4!15!}}{\dfrac{20!}{5!15!}} = \frac{19!}{4!15!} \cdot \frac{5!15!}{20!} = \frac{19!}{4!} \cdot \frac{5!}{20!} = \frac{19!}{4!} \cdot \frac{5 \cdot 4!}{20 \cdot 19!} = \frac{5}{20} = \frac{1}{4}$$ There is an easier way to see this. There are $20$ balls, of which $5$ are black and $15$ are white. If we draw a ball at random and place it in the third position, $5$ of the $20$ balls we could place there are black. Hence, the probability that a black ball is in the third position is $$\frac{5}{20} = \frac{1}{4}$$