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A loaded die has probabilities 1/21, 2/21,3/21, 4/21, 5/21, 6/21 of showing 1,2,3,4,5,6.

a) What is the prob of throwing 2 3's in succession?

So, is the answer (3/21)^2?

b) What is the prob of throwing a 4 the first time and not a 4 the second time w/ a die loaded as in a

4/21*(17/21) ???

c) if two dice loaded as in a) are thrown and we know that the sum of the #'s on the faces is >= 10, What is the prob that both are 5's?

We can get 2 6's, 1 6 and 1 5 ,and 2 5's.

So, Out of 36 events, we get the probability to be 2/36 = 1/18

d) How many times must we throw a die loaded as in a, to have a prob > 1/2 of getting an ace?

I dont know what's going on here

e) A die loaded as in a), is thrown twice. What is the prob. that the number on the die is even the first time and >4 the 2nd time?

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  • $\begingroup$ I've added an answer to Part (d). $\endgroup$ – Fly by Night Apr 21 '13 at 0:31
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Your answers to part (a) and (b) are correct.

Your answer to (c) is not. You could get (5,5), (5,6), (6,5) or (6,6). There are two ways of scoring 11.

$$P(S \ge 10) = \left(\frac{5}{21}\times \frac{5}{21}\right)+\left(\frac{5}{21}\times \frac{6}{21}\right)+\left(\frac{6}{21}\times \frac{5}{21}\right)+\left(\frac{6}{21}\times \frac{6}{21}\right) = \frac{121}{441}$$

Part (d): I thought an ace was a playing card!

Part (e): The even numbers are 2, 4, and 6. The numbers greater than four are 5 and 6, so

$$\left( \frac{2}{21}+\frac{4}{21}+\frac{6}{21} \right) \times \left(\frac{5}{21}+\frac{6}{21}\right) = \frac{44}{147}$$

EDIT

I think that by an ace you mean at least one number 1.

The way to fail is to get all not-ones. The probability of not getting a one is 20/21. The probability of getting $n$ not-ones is $(20/21)^n$. If the probability of failure is $(20/21)^n$ the the probability of success is $1-(20/21)^n$. We need to solve $1-(20/21)^n > 1/2$ for $n$. Well:

\begin{array}{ccccccc} 1-\left(\frac{20}{21}\right)^{\!n} &>& \frac{1}{2} &\iff& \frac{1}{2} & > & \left(\frac{20}{21}\right)^{\!n} \\ \\ \\\ &&&\iff& \log\left(\frac{1}{2}\right) &>& n\log\left(\frac{20}{21}\right) \\ \\ \\ &&&\iff& \frac{\log(1/2)}{\log(20/21)} &<& n \end{array} Hence $n > 14.2$, meaning that $n \ge 15$. We need to roll the dice at least 15 times to have more than a 50-50 chance of getting at least a single ace.

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  • $\begingroup$ thanks, could you explain how to do parts d, e, and f? $\endgroup$ – mary Apr 20 '13 at 23:49
  • $\begingroup$ @mary In part (d): what is an ace? I thought it was a playing card. In part (f): There is no part (f)! $\endgroup$ – Fly by Night Apr 20 '13 at 23:52
  • $\begingroup$ An ace is a special card, but I dont know what's up in this case. My bad, there's no part f $\endgroup$ – mary Apr 20 '13 at 23:53
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    $\begingroup$ ace probably means 1 $\endgroup$ – Memming Apr 20 '13 at 23:53
  • $\begingroup$ Ace does mean one; this is regional/archaic! $\endgroup$ – Sharkos Apr 20 '13 at 23:56

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