Place your final solution in the blank/space provided. All work must be shown. E
ID: 3073243 • Letter: P
Question
Place your final solution in the blank/space provided. All work must be shown. Each blank is 12 points. I. Suppose that among southeast Texans, 40% own their home, 70% own a car, and 30% own both a. What percentage of southeast Texans own at least one of the two? What percentage of car-owners are also home-owners? Round your answer to two decimal places. A 4-digit PIN is selected. What is the probability that there are no repeated digits? Compute the probability of randomly drawing five cards (without replacement) from a deck and getting exactly one ace. Three marksmen have probabilities of 0.5, 0.33, and 0.25 of hitting a target with each shot. If all three marksmen fire simultaneously, calculate the probability that at least one will hit the target. Round your answer to three decimal places. How many ways can three cars and four trucks be selected from eight cars a. and eleven trucks to be tested for a safety inspection? b. A bag contains 5 white balls and 7 black balls. Two balls are chosen at random without replacement). What is the probability that one is white and the other is black? 6. T wo machines, A and B, manufacture a particular component. Here is the chart showing each machine's effectiveness: # Components 2500 500 Probability Faulty 0.04 0.05 Machine If a component chosen at random is found to be faulty, what is the probability that it was produced by machine A?Explanation / Answer
1.
a.
A: Owns Home , P(A) = 0.40
B: Owns Car , P(B) = 0.70
A B: Owns Both, P(AB) = 0.30
Atleast one: A U B, P(A U B) = P(A) + P(B) - P(AB) = 0.40 + 0.70 - 0.30 = 0.80
b.
We find the probability that a random person who owns a car (B) also owns a home (A)
P(A | B) = P(A B) / P(B)
P(A | B) = 0.30 / 0.70 = 0.4286 = 42.86%
2.
Four digit PIN : XXXX
For each digit X we have 10 options: digits 0 to 9
Total possibilities for choosing PIN: 10 x 10 x 10 x 10
Cases where digits are not repeated: 10 x 9 x 8 x 7
Probability that the selected PIN has no repeated digits = (10x9x8x7) / (104)
= 0.504
3.
Probability of getting exactly 1 ace in a draw of five cards:
Total 52 cards: 48 non-aces , 4 aces
Draw 4 non aces cards in : (48C4) = 194580 ways
Draw 1 ace in : (4C1) = 4 ways
Total possible number of ways of drawing any 5 cards = (52C5) = 2598960
Probability of getting exactly 1 ace in a draw of five cards =
= (194580 x 4) / 2598960 = 0.30
4.
Probability that atleast one will hit the target = 1 - Probability none of the marksmen will hit the target
Probability none of the marksmen will hit the target = (1-0.5) x (1-0.33) x (1-0.25) = 0.5 x 0.67 x 0.75 = 0.25125
Probability that atleast one will hit the target = 1 - 0.25125 = 0.74875 = 0.749
5.
a.
Select 3 cars from 8 cars in : (8C3) = 56 ways
Select 4 trucks from 11 trucks : (11C4) = 330 ways
Select 3 cars and 4 trucks from 8 cars and 11 trucks in 56 x 330 = 18480 ways
b.
Bag contains: 5 white balls , 7 black balls
Probability that 1 is white and other is black.
P(Select white ball in first draw) = 5/12 , 4 white and 7 black balls remain
P(Select black ball in second draw) = 7/11
OR
P(Select black ball in first draw) = 7/12
P(select white ball in second draw) = 5/11
Total probability = (5/12)(7/11) + (7/12)(5/11) = 0.53
6.
#Compponents by Machine A = 2500
#Components by Machine B = 1500
Total #Components = 4000
P(A) = 2500/4000 = 0.625
P(B) = 1500/4000 = 0.375
Event F : Component is faulty
Given : P(F | A) = 0.04
P(F | B) = 0.05
To find: P(A | F)
By conditional law of probability:
P(A | F) = P(A F) / P(F)
P(A F) = P(A).P(F | A) = 0.025
By Total Law of probability:
P(F) = P(A).P(F | A) + P(B).P(F | B) = (0.625)(0.04) + (0.375)(0.05) =0.04375
P(A | F) = 0.025 / 0.04375 = 0.57