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STT441 SOLUTION NAME ID
- The distribution of X is given by x 0 1 2 P(X=x) 9/16 3/8 1/
a. Find Var( X ) (a) 6/8 (b) 5/8 (c) 4/8 (d) 3/
b. Find E ( 2 X ) (a) 27/16 (b) 15/816 (c) 25/16 (d) 23/
Solutions. a. E ( X )= 0 *( 9 / 16 ) 1 *( 3 / 8 ) 2 *( 1 / 16 )=1/
. E ( X^2 ) 0 *( 9 / 16 ) 1 *( 3 / 8 ) 4 *( 1 / 16 ) 5 / 8
Var ( X ) 5 / 8 ( 1 / 2 )^2 3 / 8
b. E ( 2 X ) 1 * ( 9 / 16 ) 2 *( 3 / 8 ) 4 *( 1 / 16 ) 25 / 16
- In how many ways can we arrange 6 kids in a row a. if there are no restrictions? (a) 650 (b) 660 (c) 750 (d) 720
b. if Jim and Bob must stand next to each other? (a) 150 (b) 240 (c) 250 (d) 220
c. if there are 3 pairs of friend who insist of standing next to each other? (a) 45 (b) 44 (c) 48 (d) 42
d. if there are 3 boys and 3 girls and no boys or girls are allowed to stand next to each other? (a) 45 (b) 72 (c) 78 (d) 42 Solutions a. 6!=
b. 254!=
c. 3 !* 23 48
d. 23!3!=
- Given P ( A ). 5 , P ( B ). 3 , P ( A B ). 21. Find
a. P ( Ac^ B ) (a) .11 (b) .15 (c) .13 (d).
b. P ( A Bc ) (a) .91 (b) .85 (c) .53 (d).
Solution:
A Ac
B .21 .09. c B^ .29^ .41^. .5.
So a. P ( Ac^ B )=.
b. P ( A Bc )=.21+.29+.41=.
- Find P (nomatchingbirthdays)in a group of 15 kids? (a) .75 (b) .71 (c) .63 (d).
Solution:. 75 365
- Here is some information about the events Ai , 1 i 4
4
1
i
i
i j k
i j
i
P A
P A A A i j k
P A A i j
P A i
Find ( )
4
1
i
P Ai
(a) 9/13 (b) 3/8 (c) 7/8 (d) 5/
(a) .030 (b) .006 (c) .001 (d).
c. What is the probability that a defective part was produced on one of the “good” days? (a) .23 (b) .26 (c) .31 (d).
d. What is the probability that a defective part was produced on one of the “bad” days? (a) .77 (b) .74 (c) .61 (d).
e. Two parts were produced on the same day. The first part turned out to be defective. What is the probability that the 2nd^ part is defective as well? Hint : The probability that those parts were produced on good/bad days has changed from the original setup due to the information about the 1st^ part being defective ( as you saw in the previous 2 parts). (a) .034 (b) .024 (c) .041 (d).
Solutions a. Let B denote production on bad days ( i.e. F and M) and G production on good days ( T, W, Th). We have P ( B ) 2 / 5 , P ( G ) 3 / 5. We also have P ( D | B ). 05 , P ( D | G ). 01. So : P ( D ) P ( D | B ) P ( B ) P ( D | G ) P ( G ). 05 *. 4 . 01 *. 6 . 026
b. P ( D G ) P ( D | G ) P ( G ). 01 *. 6 . 006
c. P ( G | D ) P ( G D )/ P ( D ). 006 /. 026 23 %
d. P ( B | D ) 1 P ( G | D ) 1 . 23 77 %
e. We again use the formula P ( D ) P ( D | B ) P ( B ) P ( D | G ) P ( G ) However here P ( B ). 77 , P ( G ). 23 as we found in previous 2 parts. So we get P ( D ). 05 *. 77 . 01 *. 23 . 041
- An urn contains 6 white balls and 4 black balls. A sample of 2 balls, without replacement, is being taken. Let X denotes the number of black balls in the sample. a. Find the distribution of X.
(a)
x 0 1 2 P(X=x) 8/15 1/3 2/
(b)
x 0 1 2 P(X=x) 1/3 8/15 2/
(c)
x 0 1 2 P(X=x) 2/15 1/3 8/
(d)
x 0 1 2 P(X=x) 2/15 8/15 1/
Solution:
P(X=0) = 3
, P(X=1) = 15
, P(X=2) = 15
- In a certain exam in Probability theory there was a multiple choice part that has 3 problems. Each problem has 4 possible answers. A clever student decided to use guessing rather than spending too much time thinking about the problems ( It was Friday afternoon and he had too much to do in the weekend… ). Let X be the number of answers that he got right.
a. The distribution of X is: (a) Bernoulli (.25) (b) Binomial ( 4, .5) (c) Binomial ( 4, .25) (d) Binomial( 3, .25)
b. P(X=2) (a).04 (b) .42 (c).37 (d).
Solution: b. (. 25 ) (. 75 ). 14 2
( 2 ) ^2
P X
