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There are two parts to a problem:

Part (a); In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win the jackpot, the numbers on your ticket must match the three white balls and the SuperBall. (You don't need to match the white balls in order).

If you buy a ticket, what is your probability of winning the jackpot?

Part (b):

In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win a prize, the numbers on your ticket must match at least two of the white balls or must match the SuperBall.

If you buy a ticket, what is your probability of winning a prize?

Approach for part (a):

Here is how I am solving the problem.Since 3 number in white balls needs to match during the first pick, the probability is 3/20. For the second one it will be 2/19 and 3rd one 1/18. That combined with probability for blue ball will be 3/20 * 2/19 * 1/18 * 1/10

part (b):

3 cases (1) only blue ball matches and no match in white balls (2) no match in blue ball but 2 out of 3 white ball matches (3) Both matches

Probability for case (a): 1/10 * 17/20 * 16/19 * 15/18 case (b) : 9/10 * 3/20 * 2/19 case (c) : 1/10 * 3/20 * 2/19 * 1/18

Add case (a) + (b) + (c) because they are mutually exclusive.

Am I correct in my approach?

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Ad part a) You are right.

Ad part b)

I think you have forgotten some combination. I have written down a table of combinations of balls which matches.

$\begin{array}{|c|c|} \hline \text{blue} & \text{number of white balls} \\ \hline1&2\\ \hline 1&3 \\ \hline 1&0 \\ \hline 1&1 \\ \hline 0&2 \\ \hline 0&3 \\ \hline \end{array}$

But you can also calculate the probability by using events.

A: 3 matching white balls are drawn.

B: 2 matching white balls are drawn.

C: The right blue ball is drawn.

Therefore $P(A\cup B \cup C )=P(A)+P(B)+P(C)-P(A \cap C)-P(B \cap C)$

$3/20 * 2/19 * 1/18+3* 3/20 * 2/19 * 17/18+1/10-1/10*3/20 * 2/19 * 1/18-1/10*3* 3/20 * 2/19 * 17/18\approx 14.1\%$

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Part (a) is fine !

Part (b) can be solved using only 2 cases:

(i) succeed with white balls, fail with blue ball [ best to use combinations here ]

(ii) succeed with blue ball

$$\text{Indicated Pr = }\left[\dfrac{{3\choose 2}{17\choose 1} + {3\choose 3}}{20\choose 3} \times\dfrac{9}{10}\right] + \dfrac{1}{10} =\approx 0.1411$$

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Part a is correct. For part b, case b, you are computing the chance that the first two white balls match (and the blue ball matches). You should compute the chance that exactly two of the white ball match, but that the missing one can come at any position.

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