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# For one toss of a certain coin, the probability that the

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Intern
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For one toss of a certain coin, the probability that the [#permalink]

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10 Sep 2005, 13:54
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For one toss of a certain coin, the probability that the outcome is head 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
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10 Sep 2005, 14:17
Hmm, is it E?
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10 Sep 2005, 15:43
Probability of heads at least 4 times = (Probability of heads 4 times + probability of heads 5 times).

Using binomial distribution where p=0.6 and n=5:

Probability of heads 4 times in 5 tosses:
=C(5,4)*(0.6)^4*(1-0.6)^(5-4)
=5*(0.6)^4*(0.4)^1
=5*(3/5)^4*2/5
=5*(81/625)*(2/5)
=(405/625)*(2/5)
=810/3125 = 162/625

Probability of heads 5 times in 5 tosses:
=C(5,5)*(0.6)^5*(1-0.6)^(5-5)
=1*((3/5)^5)*1
=(3/5)^5 = 243/3125

Probability of getting at least 4 heads:
= 162/625 + 243/3125
= 1053/3125 = ~1/3

Edit: I guess I didn't need to calculate that out, the answer choices go back a couple of steps. It's E.
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10 Sep 2005, 16:18
I don't understand why you have C(5,5) * (0.6)^5 for the probability that we have 5 heads.

My understanding is that to have 5 heads you multiply 5 times the probability to have a head thus = 0.6*0.6*0.6*0.6*0.6*0.6 = (0.6)^5

So for 4 heads and 1 tail we have = 0.6*0.6*0.6*0.6*0.6*(1-0.6) = (0.6)^4 * (0.4). I don't have the 5* .... can you correct me please ?
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10 Sep 2005, 17:30
Zoom, the reason we have C(5,5) is because you have 5 different ways heads can come up...

H H H H T
H T H H H
H H T H H
H H H T H
T H H H H

Get it? =)

zoom wrote:
I don't understand why you have C(5,5) * (0.6)^5 for the probability that we have 5 heads.

My understanding is that to have 5 heads you multiply 5 times the probability to have a head thus = 0.6*0.6*0.6*0.6*0.6*0.6 = (0.6)^5

So for 4 heads and 1 tail we have = 0.6*0.6*0.6*0.6*0.6*(1-0.6) = (0.6)^4 * (0.4). I don't have the 5* .... can you correct me please ?
10 Sep 2005, 17:30
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