When a random experiment is conducted,

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When a random experiment is conducted,

by gmatter2012 » Fri Jun 08, 2012 5:43 am
When a random experiment is conducted, the probability that event A occurs is 0.3. If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?
A.5/243
B.25/243
C.64/243
D.80/243
E.16/27

I think this should be

[.3 * .3 * .7 * .7 * .7 ] * 5!/ 3! 2!=

.03087 * 10 = .3087 how ever this does not match any given options
Source: — Problem Solving |

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by gmatter2012 » Fri Jun 08, 2012 5:59 am
Hi

The source had a typo

its not 0.3 but 1/3 using this 1/3 * 1/3 *2/3 * 2/3 * 2/3 * 5!/2!3!= 80/243 = D

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by Anurag@Gurome » Fri Jun 08, 2012 6:05 am
gmatter2012 wrote:When a random experiment is conducted, the probability that event A occurs is 0.3. If the random experiment is conducted 5 independent times, what is the probability that event A occurs exactly twice?
That should be 1/3.

The, the required probability = [(1/3)^2]*[(2/3)^3]*[5C2] = (8/243)*10 = 80/243 --> Option D
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by bubbliiiiiiii » Sun Jun 10, 2012 11:21 pm
Hi,

Why are we multiplying [(1/3)^2]*[(2/3)^3] by [5C2]?
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by gmat_and_me » Mon Jun 11, 2012 4:31 am
1/3 can be viewed as 1 outcome out of 3 total outcomes. So,

5 independent experiments => (3)^5 total possibilities = 243

One of the ways we can get A exactly twice in 5 experiments is

A A X X X where X stands for two other outcomes other than A. The
number of ways this arrangement can happen is

1 * 1 * 2 * 2 * 2 = 8

Another arrangement could be A X A X X. There are 5C2 such arrangements.
Hence probability is (5C2 * 8)/243 = 80/243

This is how you arrive at the solution. You may memorize the formula once
you understand the concept.

HTH,

bubbliiiiiiii wrote:Hi,

Why are we multiplying [(1/3)^2]*[(2/3)^3] by [5C2]?