Question

a) If 6n tickets numbered 1 ,1,0 ,2 K, 6n − are placed in a bag and three are drawn out,
show that the probability that the sum of the numbers on them is equal to 6n is
3n /(6n − 6()1 n − )2 . (5)
b) From a bag containing 3 white and 5 black balls, 4 balls are transferred into an empty
vessel. From this vessel a ball is drawn and is found to be white. What is the
probability that out of four balls transferred 3 are white and 1 is black?

Answer 1

Answer on Question #60646 – Math – Statistics and Probability

Question

a) If 6n tickets numbered 1, 1, 0, 2, K, 6n – are placed in a bag and three are drawn out, show that the probability that the sum of the numbers on them is equal to 6n is $3n / (6n - 1)(6n - 2)$.

Solution

First, note that 3 tickets can be drawn out from 6n tickets in $\binom{6n}{3}$ different ways. Hence,

$$n(S) = \binom{6n}{3} = \frac{6n(6n - 1)(6n - 2)}{3 \cdot 2 \cdot 1} = n(6n - 1)(6n - 2)$$

Next, let E denote the event that the sum of the numbers on the three randomly chosen tickets is 6n.

If 0 is the smallest of the three selected numbers, then there are 3n-1 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 0 & 1 & 6n - 1 \\ \vdots & \ddots & \vdots \\ 0 & 3n - 1 & 3n + 1 \end{array} \right]$$

If 1 is the smallest of the three selected numbers, then there are 3n-2 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 1 & 2 & 6n - 3 \\ \vdots & \ddots & \vdots \\ 1 & 3n - 1 & 3n \end{array} \right]$$

If 2 is the smallest of the three selected numbers, then there are 3n-4 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 2 & 3 & 6n - 5 \\ \vdots & \ddots & \vdots \\ 2 & 3n - 2 & 3n \end{array} \right]$$

If 3 is the smallest of the three selected numbers, then there are 3n-5 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 3 & 4 & 6n - 7 \\ \vdots & \ddots & \vdots \\ 3 & 3n - 2 & 3n - 1 \end{array} \right]$$

This process can be continued successively.

If 2n-4 is the smallest of the three selected numbers, then there are 5 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 2n - 4 & 2n - 3 & 2n + 7 \\ 2n - 4 & 2n - 2 & 2n + 6 \\ 2n - 4 & 2n - 1 & 2n + 5 \\ 2n - 4 & 2n & 2n + 4 \\ 2n - 4 & 2n + 1 & 2n + 3 \end{array} \right]$$

If 2n-3 is the smallest of the three selected numbers, then there are 4 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 2n - 3 & 2n - 2 & 2n + 5 \\ 2n - 3 & 2n - 1 & 2n + 4 \\ 2n - 3 & 2n & 2n + 3 \\ 2n - 3 & 2n + 1 & 2n + 2 \end{array} \right]$$

If 2n-2 is the smallest of the three selected numbers, then there are 2 possible sets of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 2n - 2 & 2n - 1 & 2n + 3 \\ 2n - 2 & 2n & 2n + 2 \end{array} \right]$$

If 2n-1 is the smallest of the three selected numbers, then there is 1 possible set of triplets each having total equal to 6n.

$$\left[ \begin{array}{ccc} 2n - 2 & 2n - 1 & 2n + 3 \\ 2n - 2 & 2n & 2n + 2 \end{array} \right]$$ $$\left[ \begin{array}{ccc} 2n - 1 & 2n & 2n + 1 \end{array} \right]$$

Since the cases listed above are mutually exclusive, it follows that

$$\begin{aligned} n(E) &= 1 + 2 + 4 + 5 + \cdots + (3n - 5) + (3n - 4) + (3n - 2) + (3n - 1) \\ &= \left[1 + 4 + \cdots + (3n - 5) + (3n - 2)\right] + \left[2 + 5 + \cdots + (3n - 4) + (3n - 1)\right] \\ &= \frac{n}{2} \left[2 + 3(n - 1)\right] + \frac{n}{2} \left[4 + 3(n - 1)\right] = \frac{n}{2} \left[6 + 6(n - 1)\right] = 3n^2 \end{aligned}$$

Thus,

$$P(E) = \frac{n(E)}{n(S)} = \frac{3n^2}{n(6n - 1)(6n - 2)} = \frac{3n}{(6n - 1)(6n - 2)}$$

Question

b) From a bag containing 3 white and 5 black balls, 4 balls are transferred into an empty vessel. From this vessel a ball is drawn and is found to be white. What is the probability that out of four balls transferred 3 are white and 1 is black?

Solution

$$P(3W1B) = \frac{\binom{3}{3}\binom{5}{1}}{\binom{3+5}{4}} = \frac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}} = \frac{\frac{3}{3!}(3 - 3)! \cdot \frac{5!}{1! (5 - 1)!}}{\frac{8!}{4! (8 - 4)!}} = \frac{1 \cdot 5}{\frac{8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4}} = \frac{1}{14}.$$

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