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

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

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New post 23 Jul 2008, 02:07
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E

Difficulty:

  55% (hard)

Question Stats:

65% (00:51) correct 35% (01:09) wrong based on 168 sessions

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For one toss of a certain coin, the probability that the outcome is heads is o.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
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-one-toss-of-a-certain-coin-the-probability-that-the-out-135559.html
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 02:12
1
tarek99 wrote:
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) \(2(.6)^4\)

c) \(3(0.6)^4 (0.4)\)

d) \(4(0.6) (0.4) + (0.6)^5\)

e) \(5(0.6)^4 (0.4) + (0.6)^5\)

please explain


4H1T + 5H

(HHHHT + HHHTH + HHTHH + HTHHH + THHHH) + (HHHHH)
5 * 0.6^4 * 0.4 + 0.6^5
Option E
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 03:46
E for me as well, using binomial theorem
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 16:45
Hello, can you tell me why you are multiplying it by 5?

5 * 0.6^4 * 0.4 + 0.6^5

Thanks.

durgesh79 wrote:
tarek99 wrote:
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) \(2(.6)^4\)

c) \(3(0.6)^4 (0.4)\)

d) \(4(0.6) (0.4) + (0.6)^5\)

e) \(5(0.6)^4 (0.4) + (0.6)^5\)

please explain


4H1T + 5H

(HHHHT + HHHTH + HHTHH + HTHHH + THHHH) + (HHHHH)
5 * 0.6^4 * 0.4 + 0.6^5
Option E
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 17:29
0.6^5 + 5C4 * (0.6^4) * (0.4)

= 0.6^5 + 5 * (0.6^4) * 0.4

E
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 18:04
Choices with 4 heads & 1 tail: 5C4 = 5 = 5*0.4*0.6\(4\)
Choice with all heads: 0.6\(5\)
0.6\(5\) + 5*0.4*0.6\(4\)
E
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 23 Jul 2008, 19:37
so you have to multiply by the #different combinations even though order doesn't matter??
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 24 Jul 2008, 03:06
pmenon wrote:
E for me as well, using binomial theorem


could you expand on that please?

Thanx
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 24 Jul 2008, 04:46
Thank you for your explanations. Makes sense now.
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 24 Jul 2008, 15:10
1
Explanation of general formula for those who asked:

Imagine that we have an experiment when the coin is tossed n times. Probability of heads in each toss is p, probability of tails is q (clearly q+p=1).

Then, the probability that from n tosses, there will be exactly k heads can be calculated as

P(k heads from n tosses) = C(k,n)*p^k*q^(n-k)

(can be proved from the reasoning that 1^n=(p+q)^n=p^n+C(1,n)*p^(n-1)*q + ... +C(n-1,n)*p*q^(n-1)+q^n - the binomial coefficients are there, thus the name)

In this problem, we need to calculate P(4 from 5)+ P(5 from 5) = [use the formula] = 5*(0.6)^4*0.4+1*(0.6)^5, which is E.

You might check wiki for 'binomial distribution' or 'binomial theorem' as well.
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 24 Jul 2008, 17:52
Alternatively,

Let X = number of times it turns up Heads (w.p. 0.6)

X~Binomial(5,0.6)

Pr (X>=4)
= Pr(X=4) + Pr(X=5)
= (5C4)(0.6)^4 0.4 + (5C5)0.6^5

which is E.
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 24 Jul 2008, 18:51
1
greenoak wrote:
Explanation of general formula for those who asked:

Imagine that we have an experiment when the coin is tossed n times. Probability of heads in each toss is p, probability of tails is q (clearly q+p=1).

Then, the probability that from n tosses, there will be exactly k heads can be calculated as

P(k heads from n tosses) = C(k,n)*p^k*q^(n-k)

(can be proved from the reasoning that 1^n=(p+q)^n=p^n+C(1,n)*p^(n-1)*q + ... +C(n-1,n)*p*q^(n-1)+q^n - the binomial coefficients are there, thus the name)

In this problem, we need to calculate P(4 from 5)+ P(5 from 5) = [use the formula] = 5*(0.6)^4*0.4+1*(0.6)^5, which is E.

You might check wiki for 'binomial distribution' or 'binomial theorem' as well.


Thanks. It looks complicated. I will stick to my approach
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Re: For one toss of a certain coin, the probability that the out  [#permalink]

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New post 29 Jul 2014, 00:11
tarek99 wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.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
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5


\(P(h)=0.6\), so \(P(t)=0.4\). We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: \(P(h=5)=0.6^5\);

4h and 1t: \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, \(P(h\geq{4})=0.6^5+5*0.6^4*0.4\).

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-one-toss-of-a-certain-coin-the-probability-that-the-out-135559.html
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