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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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I've added an answer to Part (d). –  Fly by Night Apr 21 '13 at 0:31

1 Answer 1

up vote 2 down vote accepted

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

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