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Selecting the color blind person in no more than 3 tries

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Selecting the color blind person in no more than 3 tries

by gmattesttaker2 » Tue May 27, 2014 6:32 pm
Hello,

For the following:

A randomly selected sample population consists of 60% women and 40% men. 90% of the women and 15% of the men are colorblind. For a certain experiment, scientists will select one person at a time until they have a colorblind subject. What is the approximate probability of selecting a colorblind person in no more than three tries?

A) 95%
B) 90%
C) 80%
D) 75%
E) 60%

OA: [spoiler]95%[/spoiler]


I was able to come up with the double set matrix but then couldn't solve for the probability part. This is what I have:

60/100 + (40/100).(60/99) + (40/100)(39/99)(60/98)

I wasn't sure though how to simplify this further. Thanks for your help.

Regards,
Sri
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by [email protected] » Tue May 27, 2014 9:38 pm
Hi Sri,

There are a couple of different ways to solve this problem. Here's how I solved it.

Based on the given info, we have....

60% Women
40% Men

Color-Blind:
90% of Women = (.9)(60%) = 54% of total group
15% of Men = (.15)(40%) = 6% of the total group

NOT Color-Blind = everyone else = 40% of the total group

Probability Rule:

Want + Don't Want = 1

When a probability question gives us multiple tries to accomplish 1 task, it's usually fastest to calculate the probability that we DO NOT accomplish the task, then subtract that from the number 1.

Instead of calculating the probability of picking a color-blind person in 3 tries, we'll calculate the probability that we DO NOT select a color-blind person in 3 tries....

Notice the word "approximate"....we can keep things simple....

(NOT)(NOT)(NOT) = (.4)(.4)(.4) = .064 This is the probability of NOT selecting a color-blind person in 3 tries.

The probability of selecting a color-blind person in 3 tries is....

1 - .064 = .936

This is really close to....

Final Answer: A

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by buddyme » Tue Sep 22, 2015 1:32 am
[email protected] wrote:Hi Sri,

There are a couple of different ways to solve this problem. Here's how I solved it.

Based on the given info, we have....

60% Women
40% Men

Color-Blind:
90% of Women = (.9)(60%) = 54% of total group
15% of Men = (.15)(40%) = 6% of the total group

NOT Color-Blind = everyone else = 40% of the total group

Probability Rule:

Want + Don't Want = 1

When a probability question gives us multiple tries to accomplish 1 task, it's usually fastest to calculate the probability that we DO NOT accomplish the task, then subtract that from the number 1.

Instead of calculating the probability of picking a color-blind person in 3 tries, we'll calculate the probability that we DO NOT select a color-blind person in 3 tries....

Notice the word "approximate"....we can keep things simple....

(NOT)(NOT)(NOT) = (.4)(.4)(.4) = .064 This is the probability of NOT selecting a color-blind person in 3 tries.

The probability of selecting a color-blind person in 3 tries is....

1 - .064 = .936

This is really close to....

Final Answer: A

GMAT assassins aren't born, they're made,
Rich
Hi Rich, Is 0.936 the exact answer or is it an approximaation?
My doubt is that even if you have to calculate the probability that we DO NOT select a color-blind person in 3 tries, according to my undestanding, it hs to befollowing

40/100 * 39/99 * 38/98 and not 0.4 * 0.4 * 0.4 (since the selected candidate will not be placed back again)
If this is wrong, can you explain why is it wrong?

Actually if i take the total sample to be 200 instead of 100 then my answer will be different according to my calculations. So I am suspecting that the question itself is incorrect
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by [email protected] » Tue Sep 22, 2015 8:24 am
Hi buddyme,

Notice the wording of the question: "what is the APPROXIMATE probability..." - this means that whatever calculation you do will probably not be an exact match for any of the answer choices, so some 'rounding' will be required.

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by Matt@VeritasPrep » Thu Sep 24, 2015 10:22 am
Let's call our population x.

.9*.6x + .4*.15x are colorblind, so .6x are colorblind.

The probability of selecting a colorblind person -- I assume they mean "at least one" colorblind person -- is just 1 - (probability of NO colorblind people), which is easier to compute.

Prob(no colorblind) = (first not colorblind) * (second not colorblind) * (third not colorblind)

They don't specify whether we're randomly selecting three different people (another big ambiguity), but let's assume that's NOT the case. We thus have

Prob(no colorblind) = .4x * .4x * .4x = .064x

So our probability of selecting a colorblind person is x - .064x, or .936x. No approximation required ... but the question is so ambiguously worded that we can't be perfectly sure 93.6% is the exact answer.
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