1-P method doesnt work.Please help

Expert replies

dddanny2006: Master | Next Rank: 500 Posts; Posts: 209; Joined: Thu Jan 12, 2012 12:59 pm

by dddanny2006 » Mon Nov 25, 2013 11:14 am

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

Method 1

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

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

There 1-P=0.52

Why am I wrong here?
Answer is 0.68

Thanks

Dan

Source: Beat The GMAT — Problem Solving | Mon Nov 25, 2013 11:14 am

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by [email protected] » Mon Nov 25, 2013 3:09 pm

Hi dddanny2006,

In this question, we're told that the probability of choosing a car in a certain color is:

.09 = green
.15 = white
.21 = red
.23 = blue
.32 = some other color (this is implied, since total probability always adds up to 1)

We're asked for the probability of picking a car in ANY of those 4 colors (meaning green OR white OR red OR blue).

The probability formula = (What we want) / (Total possibilities)

Since we want "green or white or red or blue", we can simply add those up: .09 + .15 + .21 + .23 = .68

.68/1 = .68

The probability of NOT getting one of those colors is .32/1 = .32
So, 1 = .32 = .68

You can get the same answer with either approach, as long as you're doing the calculation that the question asks for.

GMAT assassins aren't born, they're made,
Rich

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dddanny2006: Master | Next Rank: 500 Posts; Posts: 209; Joined: Thu Jan 12, 2012 12:59 pm

by dddanny2006 » Mon Nov 25, 2013 3:21 pm

What wrong did I do in the question,or in other words what should have been the question for the answer I got via my method

Thanks

[email protected] wrote:Hi dddanny2006,

In this question, we're told that the probability of choosing a car in a certain color is:

.09 = green
.15 = white
.21 = red
.23 = blue
.32 = some other color (this is implied, since total probability always adds up to 1)

We're asked for the probability of picking a car in ANY of those 4 colors (meaning green OR white OR red OR blue).

The probability formula = (What we want) / (Total possibilities)

Since we want "green or white or red or blue", we can simply add those up: .09 + .15 + .21 + .23 = .68

.68/1 = .68

The probability of NOT getting one of those colors is .32/1 = .32
So, 1 = .32 = .68

You can get the same answer with either approach, as long as you're doing the calculation that the question asks for.

GMAT assassins aren't born, they're made,
Rich

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