Probabality-Similar questions,methods contradict the answer.

[dddanny2006]

If the probabilities are respectively 0.09,0.15,.21 and 0.23 that a person purchasing a new automobile will choose the color green,white,red or blue,what is the probability that a given buyer will purchase a new automobile that comes in one of those colors.

Answer 0.68---------------Simple method

Source-Probability for Engineers and Scientists by Walpole.

Method 1

Prob(of buying one of those colours)= 0.09 + (.91*0.15)+(0.91*0.85*0.21)+(0.91*0.85*0.79*0.23)=0.52

But the books says 0.68,why am I wrong?

No problem,Ill do the second method Method 2

Prob(of choosing atleast one of those colours)=1-prob(of not choosing atleast one of those colours)

Prob(of not choosing one of those colours)= 0.91*0.85*0.79*0.77=0.48

There 1-P=1-0.48=0.52

But Im wrong in both cases. Why?

The answer is 0.68 ,a pretty straight forward way to get there.But I dont know why my method is wrong?

2)What is the difference between my question above and the one here below-

On his drive to work,Leo listens to one of the three radio stations a,b,or c.He first turns to A.If a is playing a song he likes,he listens to it;if not he turns to b.If b is playing a song he likes,he listens to it;if not he turns to c.If c is playing a song he likes,he listens to it;it not,he turns off the radio.For each station,the probability is 0.30 that at any given moment the station is playing a song he likes.On his drive to work,what is the probability that Leo will hear a song he likes?

a. 0.027
b. 0.090
c. 0.417
d. 0.657-----Answer
e. 0.900

I know how to solve this.I wanted to differentiate between the two problems.

3)Why is the 1-P method that Ive used above wrong?At least that should have been right.Because the probability of him choosing atleast one car is the same as probability of choosing one car of the given colours.

Thank you Ron.

[RonPurewal]

When you're looking at alternatives, you can just add the probabilities. So, 0.09 + 0.15 + 0.21 + 0.23 = 0.68.

You don't multiply probabilities together unless you are dealing with separate issues. E.g., your expression 0.91*0.85*0.79*0.77 would be the probability that one car is not green, and that a second car is not white, and that a third car is not red, and that a fourth car is not blue.

[dddanny2006]

I did not understand.0.09 is the probability that he buys a green car,implies probability of not buying the green car is 0.91 right.I dont get the point Ron :(

When you're looking at alternatives, you can just add the probabilities. So, 0.09 + 0.15 + 0.21 + 0.23 = 0.68.

You don't multiply probabilities together unless you are dealing with separate issues. E.g., your expression 0.91*0.85*0.79*0.77 would be the probability that one car is not green, and that a second car is not white, and that a third car is not red, and that a fourth car is not blue.

[RonPurewal]

I did not understand.0.09 is the probability that he buys a green car,implies probability of not buying the green car is 0.91 right.I dont get the point Ron :(

When you're looking at alternatives, you can just add the probabilities. So, 0.09 + 0.15 + 0.21 + 0.23 = 0.68.

You don't multiply probabilities together unless you are dealing with separate issues. E.g., your expression 0.91*0.85*0.79*0.77 would be the probability that one car is not green, and that a second car is not white, and that a third car is not red, and that a fourth car is not blue.

Yes, 0.91 is the probability that the first car isn't green. If you're going to multiply that by the probabilities of other colors, that implies that you're dealing with different cars.