Expert: Ahmed Salami - 8/6/2004

Question
1. When two fair dice are rolled, what is the probability of getting a 4 on the first dice and a 2 on the second?

2. On the roll og a single die, you win $5 if 1 or 2 turns up an $10 if 3 or 4 turns up. You do not win or lose if 5 turns up. How much should you lose if a 6 turns up in order for the game to be fair?

3. Calculate the range, variance, and standard deviation for each of the following samples:

a) 1, 2, 3 ,4, 5, 6

b) 8, -2, 1, 3, 5, 4, 4, 1, 3, 4, 5

4. What is the probability of getting at least one heart in a 5-card hand dealt from a standard 52-card deck?

5. Find n, given that Pn,4 = 56Pn,2

6. How many different permutations can be made of the letters of the word "dependent"?

7. How many rectangles are formed when 8 vertical lines are intersected by 5 horizontal lines?

8. Find n, given that Cn,2 = 78

Urn 1 contains 4 white balls and 3 black balls, and Urn 2 contains 1 white ball and 2 black balls. A man selects one ball from one of the vases. He chooses randomly. What is the probability that hs will select a white ball?


Answer
Hi Nathan,
Forgive me for the delay.
1)When two fair dice are rolled, the outcome on each die is independent of the other.
The probability of getting a 4 on the first die
= 1/6
The probability of getting a 2 on the second is
= 1/6
Due to the fact that they are independent events, the probability of getting a 4 on the first die and a 2 on the second is therefore 1/6 x 1/6 = 1/36

2)You could win $5 or $10 but not more. If the game should be fair, i'm of the opinion that you should not lose more than the maximum you could win which is $10.

3)I'm not actually going to do this but i will tell you how. The range is always the highest value minus the lowest value. The variance is the sum of the squares of all the differences from the mean of all the numbers divided by the amount of numbers. The standard deviation is the square root of the variance. Alright i'm going to do the first one.
The range is 6 - 1 = 5
The mean is (1+2+3+4+5+6)/6 = 21/6 = 3.5
The differences from the mean are -2.5,-1.5,-0.5,0.5,1.5,2.5
The squares of these are 6.25,2.25,0.25,0.25,2.25,6.25
The sum of these is 6.25+2.25+0.25+0.25+2.25+6.25 = 17.5
The variance is therefore 17.5/6 = 2.92
The s.d is 1.71

4)The probability of getting r successes in n trials when the probabilities of success and failure are p and q respectively is nCr.(p^r)(q^n-r)
So for at least one heart in a 52-card deck containing 13 hearts, p = 13/52 = 1/4, q = 3/4, n = 5, we now take
r = 1,2,3,4,5 in turns and evaluate as in the formula above and add the results.

5)Pn,4 = n!/(n-4)!
 56Pn,2 = 56 n!/(n-2)!
Therefore,
n!/(n-4)! = 56 n!/(n-2)!
(n-2)!/(n-4)!= 56
(n-2)(n-3)= 56
n^2 - 5n + 6 = 56
n^2 - 5n - 50 = 0
(n-10)(n+5)= 0
n = 10 or -5  but we discard -5 because it is negative
So n = 10

6)The word "dependent" has nine letters but with 2 d's,
3 e's and 2 n's.
Therefore, the number of different permutations can be made
= 9!/2!.3!.2!
= 15120

7)There are 7 spaces between 8 vetical lines and 4 spaces between 5 horizontal lines. The number of rectangles formed is therefore 7 x 4 = 28

8)Cn,2 = 78
 n!/(n-2)!.2! = 78
 n(n-1)= 78 x 2
 n^2 - n = 156
 n^2 - n - 156 = 0
 (n-13)(n+12)= 0
 n = 13, we discard -12

9)The probability of choosing from anyone of the vases is 1/2.
The probability of choosing a white ball from urn 1 = 4/7
The probability of choosing a white ball from urn 2 = 1/3
Therefore,
The probability of choosing a white ball from anyone of the vases is (1/2 x 4/7)+(1/2 x 1/3)
= 2/7 + 1/6 = 19/42

I hope you still find it helpful.
I also don't usually answer too long homework questions.
I'm sorry if i haven't really made too explanatory comments.
Good luck.
Regards.