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Probability- tough one!

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by target_mba2010 » Sat Sep 05, 2009 7:23 am
This is from GMAT Prep2-

For one toss of a certain coin,the probability that the outcome is heads is 0.6.If this coin is tossed 5 times,which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4*(0.4)
D. 4(0.6)^4*(0.4)+(0.6)^5
E. 5(0.6)^4*(0.4)+(0.6)^5

Please do reply ASAP. I am not able to crack this one!
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by krumpy » Sat Sep 05, 2009 7:39 am
I would choose E.

You need the prob of 5 times b/c that qualifies as at least four times as well as the prob that you get exactly 4 heads. That is going to be 0.6^4*0.4 (4heads x one tail). Need to multiply by 5 as there are five differet ways for this outcome to occur (HHHHT,HHHTH,HHTHH,HTHHH,THHHH).

https://en.wikipedia.org/wiki/Binomial_probability
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Re: Probability- tough one!

by maihuna » Sat Sep 05, 2009 7:53 am
target_mba2010 wrote:This is from GMAT Prep2-

For one toss of a certain coin,the probability that the outcome is heads is 0.6.If this coin is tossed 5 times,which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4*(0.4)
D. 4(0.6)^4*(0.4)+(0.6)^5
E. 5(0.6)^4*(0.4)+(0.6)^5
one!
Prob for r out of n is nCrp^rq^n-r

For 4: 5c4*0.6^4*0.4^1
For 5: 5c5*0.6^5

Total sum = 5*0.6^4*0.4 + 0.6^5 E
Charged up again to beat the beast :)
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by Nermal » Sat Sep 05, 2009 7:58 am
You have to calculate the probability of getting heads 4 times and add to it the probability of getting heads 5 times.

4 times: 5*0,6*0,6*0,6*0,6*0,4 = 5*0,6^4*0,4
since it doesn't matter when you get the tail(first toss,...., last toss)

5 times: 0,6^5
there is, of course, only one possibility

5*(0,6^4)*0,4 + 0,6^5

IMO it's E
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by target_mba2010 » Sat Sep 05, 2009 10:16 pm
The OA is E.

I got the logic.
Thanks for the responses guys.

Good Luck!
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