Probability- tough one!

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target_mba2010: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Wed Aug 12, 2009 9:01 pm

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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Source: Beat The GMAT — Problem Solving | Sat Sep 05, 2009 7:23 am

krumpy: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Mon Aug 10, 2009 7:12 am

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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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

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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Nermal: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Sun Jun 21, 2009 10:25 am; Location: Germany; Thanked: 7 times

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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target_mba2010: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Wed Aug 12, 2009 9:01 pm

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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