Math, asked by Zondertron, 12 hours ago

In a city, a person own independently a sedan car with probability 3/10 and SUV with probability 410.410. If he has sedan only, then he keeps a driver with probability 6/10, whereas if he owns SUV only, then he keeps a driver with probability 7/10, whereas if he keeps both type of cars then his probability of keeping a driver is 9/10. Then

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Answered by amitnrw
13

Given:  a person own independently a sedan car with probability 3/10 and SUV with probability 4/10.

If he has sedan only, then he keeps a driver with probability 6/10

If he has SUV only, then he keeps a driver with probability 7/10

if he keeps both type of cars then his probability of keeping a driver is 9/10.

To Find :

Probability of keeping driver

Probability that he owns SUV if he keeps a driver

Solution:

Probability Sedan car = P(A) = 3/10

Proabability SUV car = P(B) = 4/10

A and B are independent events

Hence P (A ∩ B) = P(A).P(B) = (3/10)(4/10) = 12/100

Proabability Sedan only = P(A) -P (A ∩ B)  = 3/10 - 12/100  = 18/100

Proabability SUV only =    P(B) -P (A ∩ B)  =4/10 - 12/100  = 28/100

Probability both = 12/100

probability of keeping a driver= (18/100)(6/10) + (28/100)(7/10) + (12/100)(9/10)

= (108 + 196 + 108)/1000

= 412 /1000

Probability that person keeps driver = 412/1000

probability own SUV if he keeps driver  =  ( 196 + 108 ) / 412

= 304/412

= 76/103

Probability that he owns SUV if he keeps a driver = 76/103

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