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Suppose that a box contains r red balls, w white balls, and b blue balls. Suppose also that balls are drawn from the box one at a time, at random, without replacement. What is the probability that all r red balls will be obtained before any white balls are obtained?

Short Answer

Expert verified

The probability is \(\frac{{r!w!}}{{\left( {r + w} \right)!}}\)

Step by step solution

01

Given information

The box contains r red balls, w white balls, and b blue balls.

The balls are drawn randomly without replacement from the box.

02

Calculate the probability

Therefore, the total number of balls in the box is,

\(r + w + b\)

Here, no blue balls are used, so keep them aside.

Therefore, only\(r + w\)are useful.

Therefore, all the red balls are drawn before any white balls if and only if the first r recorded draws are red.

Therefore\({}^{r + w}{C_r}\)are red.

Choices for choosing all red balls before white and only one of them is all red balls.

Therefore, the probability of all r red balls will be obtained before any white balls is,

\(\begin{aligned}{}\frac{1}{{{}^{r + w}{C_r}}} &= \frac{1}{{\frac{{\left( {r + w} \right)!}}{{r!\left( {r + w - r} \right)!}}}}\\ &= \frac{1}{{\frac{{\left( {r + w} \right)!}}{{r!w!}}}}\\ &= \frac{{r!w!}}{{\left( {r + w} \right)!}}\end{aligned}\)

Therefore, the probability is \(\frac{{r!w!}}{{\left( {r + w} \right)!}}\)

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Most popular questions from this chapter

Consider a state lottery game in which each winning combination and each ticket consists of one set of k numbers chosen from the numbers 1 to n without replacement. We shall compute the probability that the winning combination contains at least one pair of consecutive numbers.

a. Prove that if\({\bf{n < 2k - 1}}\), then every winning combination has at least one pair of consecutive numbers. For the rest of the problem, assume that\({\bf{n}} \le {\bf{2k - 1}}\).

b. Let\({{\bf{i}}_{\bf{1}}}{\bf{ < }}...{\bf{ < }}{{\bf{i}}_{\bf{k}}}\)be an arbitrary possible winning combination arranged in order from smallest to largest. For\({\bf{s = 1,}}...{\bf{,k}}\), let\({{\bf{j}}_{\bf{s}}}{\bf{ = }}{{\bf{i}}_{\bf{s}}}{\bf{ - }}\left( {{\bf{s - 1}}} \right)\). That is,

\(\begin{array}{c}{{\bf{j}}_{\bf{1}}}{\bf{ = }}{{\bf{i}}_{\bf{1}}}\\{{\bf{j}}_{\bf{2}}}{\bf{ = }}{{\bf{i}}_{\bf{2}}}{\bf{ - 1}}\\{\bf{.}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{j}}_{\bf{k}}}{\bf{ = }}{{\bf{i}}_{\bf{k}}}{\bf{ - }}\left( {{\bf{k - 1}}} \right)\end{array}\)

Prove that\(\left( {{{\bf{i}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{i}}_{\bf{k}}}} \right)\)contains at least one pair of consecutive numbers if and only if\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)contains repeated numbers.

c. Prove that\({\bf{1}} \le {{\bf{j}}_{\bf{1}}} \le ... \le {{\bf{j}}_{\bf{k}}} \le {\bf{n - k + 1}}\)and that the number of\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)sets with no repeats is\(\left( {\begin{array}{*{20}{c}}{{\bf{n - k + 1}}}\\{\bf{k}}\end{array}} \right)\)

d. Find the probability that there is no pair of consecutive numbers in the winning combination.

e. Find the probability of at least one pair of consecutive numbers in the winning combination

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