# Discrete probability distribution

• Jul 6th 2011, 08:05 AM
JWEI
Discrete probability distribution
A hotel gets cars for its guests from three rental agencies, 34% agency A, 46% agency B,20% agency C.If 5% of the cars from A, 8% of the cars from B, and 10% of the cars from C need tune-ups, find probability
i) a car needing a tune-up will be delivered to one of the guess.
ii) if a car needing a tune up is delivered to one of the guests, it came from agency B.

who can help me solve this question. thanks ><
• Jul 6th 2011, 10:56 AM
waqarhaider
Re: Probability Question
From given information 1.7 % from A , 3.68 % from B and 2% from C need tuning. So total 7.38 % , therefore

(i) Probability = 0.00738 and

(ii) Probability = 0.00368
• Jul 6th 2011, 11:51 AM
SpringFan25
Re: Probability Question
Quote:

Originally Posted by waqarhaider
From given information 1.7 % from A , 3.68 % from B and 2% from C need tuning. So total 7.38 % , therefore

(i) Probability = 0.00738 and

(ii) Probability = 0.00368

Im not sure those answers are correct.

Quote:

Originally Posted by JWEI
A hotel gets cars for its guests from three rental agencies, 34% agency A, 46% agency B,20% agency C.If 5% of the cars from A, 8% of the cars from B, and 10% of the cars from C need tune-ups, find probability
i) a car needing a tune-up will be delivered to one of the guess.
ii) if a car needing a tune up is delivered to one of the guests, it came from agency B.

who can help me solve this question. thanks ><

Events:
A = Car from garage A
B = Car from garage B
C = Car from garage C
T = Car Needs tune up

i) a car needing a tune-up will be delivered to one of the guests.
$\displaystyle P(T) = P(A \cap T) + P(B \cap T) + P(C \cap T)$
$\displaystyle P(T) = (0.34*0.05) + (0.46 * 0.08) + (0.2*0.1) = 0.0738$ (note only 1 zero after the decimal)

ii) if a car needing a tune up is delivered to one of the guests, it came from agency B.
use the formula for conditional probability
$\displaystyle P(B|T) = \frac{P(B \cap T)}{P(T)}$

$\displaystyle P(B|T) = \frac{0.46 * 0.08}{0.0738} = 0.4985$
• Jul 6th 2011, 12:21 PM
JWEI
Re: Probability Question
Quote:

Originally Posted by SpringFan25
Im not sure those answers are correct.

Events:
A = Car from garage A
B = Car from garage B
C = Car from garage C
T = Car Needs tune up

i) a car needing a tune-up will be delivered to one of the guests.
$\displaystyle P(T) = P(A \cap T) + P(B \cap T) + P(C \cap T)$
$\displaystyle P(T) = (0.34*0.05) + (0.46 * 0.08) + (0.2*0.1) = 0.0738$ (note only 1 zero after the decimal)

ii) if a car needing a tune up is delivered to one of the guests, it came from agency B.
use the formula for conditional probability
$\displaystyle P(B|T) = \frac{P(B \cap T)}{P(T)}$

$\displaystyle P(B|T) = \frac{0.46 * 0.08}{0.0738} = 0.4985$

SpringFan25~ Thx for your help!! ^^