SOLUTION: A radio station has a contest in which contestants toll a regular 6-sided die. If he rolls a 1 or 2, he wins $50. If he rolls a 3 or a 4, he wins $100. If he rolls a 5, he wins $10

Algebra ->  Probability-and-statistics -> SOLUTION: A radio station has a contest in which contestants toll a regular 6-sided die. If he rolls a 1 or 2, he wins $50. If he rolls a 3 or a 4, he wins $100. If he rolls a 5, he wins $10      Log On


   

Question 1152116: A radio station has a contest in which contestants toll a regular 6-sided die. If he rolls a 1 or 2, he wins $50. If he rolls a 3 or a 4, he wins $100. If he rolls a 5, he wins $1000. If he rolls a 6, he doesn't win anything. What is the probability that out of the first 5 contestants exactly 2 win $100?
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


P(win $100) = 2/6 = 1/3 [win $100 with either a 3 or a 4]
P(other outcome ("lose")) = 2/3

The "probability vector" for each roll is

%281%2F3%29W%2B%282%2F3%29L [1/3 chance of "winning" (W); 2/3 chance of "losing" (L)]

The probability that exactly 2 out of 5 contestants win $100 is the coefficient of the "(W^2)(L^3)" term in the expansion of %28%281%2F3%29W%2B%282%2F3%29L%29%5E5

Expand using the binomial theorem. The coefficient of the (W^2)(Y^3) term is

= 0.3292181 to 7 decimal places.

If you have a TI83 calculator, you can confirm that result with

2nd-VARS
binompdf(5,1/3,2) [5 trials; 1/3 probability of winning; 2 successes]