Determining the Expected Draw Before Picking a Red Ball from a Box
The problem of determining the expected number of white balls to be picked before encountering a red ball can be approached in various ways. This article explores both direct and distribution-based methods to arrive at the expected number of draws.
Direct Expectation Approach
Let's begin with a straightforward approach. We start by considering the probability of drawing red balls before encountering a white ball. Since there are ( w ) white balls and ( r ) red balls in a box, the probability of drawing a red ball at the first draw is ( frac{r}{w r} ).
Given that the first draw is a red ball, the probability of the second draw being red (without replacement) is ( frac{r-1}{w r-1} ). This process continues, with the probability of each subsequent red ball draw decreasing until the first white ball is drawn.
The expected number of red balls drawn before the first white ball follows a pattern. For the first draw, the probability is ( frac{8}{10} ). For the second, it is ( frac{7}{9} ), and so on. When summed up, this series converges, and we find that by the third draw, the odds of drawing a red ball and a white ball are approximately equal. Thus, the expected number of draws before a white ball is approximately 3.
On a more detailed analysis, one can calculate the exact expected number of draws up until a white ball is picked. Here are the probabilities for each draw:
On the 1st draw: ( frac{8}{10} 0.8 )
2nd draw: ( frac{7}{9} approx 0.778 )
3rd draw: ( frac{6}{8} 0.75 )
4th draw: ( frac{5}{7} approx 0.714 )
5th draw: ( frac{4}{6} 0.667 )
6th draw: ( frac{3}{5} 0.6 )
7th draw: ( frac{2}{4} 0.5 )
8th draw: ( frac{1}{3} approx 0.333 )
9th draw: ( frac{0}{2} 0 )
Adding these probabilities shows that by the 8th draw, the probability of picking a white ball is approaching 66.67%, and by the 9th draw, it is 100%. This aligns with the direct approach mentioned earlier.
Negative Hypergeometric Distribution
The negative hypergeometric distribution is another useful tool for solving such problems. The distribution describes the number of failures before a specified number of successes in a hypergeometric distribution.
In this problem, the number of successes (white balls) is 1, and we are interested in the number of failures (red balls) before the first success. The expected number of failures in the negative hypergeometric distribution is given by the formula:
[ E(X) frac{r}{p} - (m - 1) ]
where ( r ) is the number of red balls, ( w ) is the number of white balls, and ( p ) is the total number of balls. Using the values ( w 10 ) and ( r 8 ), the expected number of red balls before the first white ball is:
[ E(X) frac{8}{frac{8 10}{18}} - (8-1) frac{8}{frac{18}{18}} - 7 8 - 7 frac{8}{3} ]
This means that on average, we can expect to draw ( frac{8}{3} ) red balls before drawing the first white ball. Adding the one white ball to the expected number of red balls, the total expected number of draws is:
[ frac{8}{3} 1 frac{11}{3} approx 3.67 ]
Linearity of Expectation
Another method involves using the linearity of expectation. We can decompose the problem into the expected number of red balls and the expected number of white balls. Since we want one white ball and the expected number of red balls minus one (the one red ball that will be replaced by the white ball) is ( frac{8}{3} ), the expected number of draws is:
[ E(text{total draws}) frac{8}{3} 1 frac{11}{3} ]
This method confirms our previous result and provides an alternative perspective on solving the problem.
Conclusion
The expected number of draws before picking a white ball, whether through direct probability analysis, negative hypergeometric distribution, or linearity of expectation, converges to the value ( frac{11}{3} ). This result facilitates understanding and solving similar problems in statistics and probability.