Question 1152116
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P(win $100) = 2/6 = 1/3  [win $100 with either a 3 or a 4]
P(other outcome ("lose")) = 2/3<br>
The "probability vector" for each roll is<br>
{{{(1/3)W+(2/3)L}}}  [1/3 chance of "winning" (W); 2/3 chance of "losing" (L)]<br>
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 {{{((1/3)W+(2/3)L)^5}}}<br>
Expand using the binomial theorem.  The coefficient of the (W^2)(Y^3) term is<br>
{{{(C(5,2))*(1/3)^2*(2/3)^3 = (10)(1/9)(8/27) = 80/243}}} = 0.3292181 to 7 decimal places.<br>
If you have a TI83 calculator, you can confirm that result with<br>
2nd-VARS
binompdf(5,1/3,2) [5 trials; 1/3 probability of winning; 2 successes]<br>