Ira S. answered • 10/09/16

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The probability that you roll a 1 on a single die is 1/6 for each roll.

So if you want exactly 2 occurrences, that would be (1/6)

^{2}* (5/6)^{8}. That's a 1 twice and something else 8 times to make up all ten rolls.But we're not done. The 1 could have been on the first two rolls, the last two rolls, the third and fifth roll....we don't know.

That's why you need to multiply this answer by

_{10}C_{2}, (I hope you know this notation) which equals (10*9)/(2*1)=45.^{}So your answer is 45*(1/6)

^{2}*(5/6)^{8 }= .2907 or approximately 29%of the time.Hope this helped.

Kenneth S.

Using the Binomial Distribution Commands on Ti-83 or 84:

Commands for binomial distributions (Bernoulli Trials), p = PROBABILITY OF SUCCESS, n TRIALS:

• binompdf (n, p) to produce the entire list of probabilities, x from 0 to n

• binomcdf (n, p, k) to get the (cumulative) probability of any x value from 0 to k.

10/09/16