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For one toss of a certain coin, the probability that the outcome is he

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For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post Updated on: 14 Oct 2019, 04:34
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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. \(3(0.6)^4\)

D. \(4(0.6)^4(0.4) + (0.6)^5\)

E. \(5(0.6)^4(0.4) + (0.6)^5\)

Originally posted by ralucaroman on 25 Oct 2009, 18:38.
Last edited by Bunuel on 14 Oct 2019, 04:34, edited 3 times in total.
Renamed the topic, edited the question added the answer choices and OA.
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 25 Oct 2009, 18:54
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3
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
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.
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 25 Oct 2009, 19:12
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Bunuel - the quant maestro --- nails one more problem... :))

I missed the multiplying by 5 too
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 19 Sep 2014, 20:00
Bunuel wrote:
Stiv 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.


Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 20 Sep 2014, 13:22
kpali wrote:
Bunuel wrote:
Stiv 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.


Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.


Think logically, the case of 4 heads and 1 tail can occur in several ways (each of which having the same probability), so we have to account for the order here.
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 31 May 2017, 23:17
1
Stiv 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(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

1. Probability of getting 4 heads and one tail is 5(0.6)^4 * (0.4). 5 is nothing but 5C4 ways. of 4 heads happening.
2. Probability of getting heads all the 5 times is (0.6)^5
3. Total probability is answer E.
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 25 Jun 2017, 12:29
2
Stiv 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(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



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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 18 Oct 2018, 18:13
1
Stiv 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. \(2(0.6)^4\)

C. \(3(0.6)^4\)

D. \(4(0.6)^4(0.4) + (0.6)^5\)

E. \(5(0.6)^4(0.4) + (0.6)^5\)


The outcome “at least 4 heads out of 5 flips” means that we can get either exactly 4 heads out of 5 flips OR 5 heads out of 5 flips.

First, we need to determine the probability of exactly 4 heads out of 5 flips, so:

P(HHHHT) = (0.6)^4 x 0.4

Since HHHHT can be arranged in 5!/4! = 5 ways, the total probability of obtaining exactly 4 heads is 5(0.6)^4 x 0.4.

The probability of 5 heads is (0.6)^5.

So the total probability is 5(0.6)^4 x 0.4 + (0.6)^5.

Answer: E
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 16 Aug 2019, 06:59
Bunuel wrote:
Stiv 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.

Plz clear my doubt:

Bunuel

if it is asked , 5 coins are tossed what is probabilty of getting 4 heads exactly
so we cn
HHHHT so
5!/4! ways /2^5(total ways) but why are we not considering 2^5 in denominator here ,plz clarify
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Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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New post 16 Aug 2019, 07:37
vanam52923 wrote:
Bunuel wrote:
Stiv 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.

Plz clear my doubt:

Bunuel

if it is asked , 5 coins are tossed what is probabilty of getting 4 heads exactly
so we cn
HHHHT so
5!/4! ways /2^5(total ways) but why are we not considering 2^5 in denominator here ,plz clarify


If it were a fair coin, so if the probability of a tail = probability of a head = 1/2, then it would be P(h = 4) = 5!/4!*(1/2)^4*1/2= 5!/4!*(1/2)^5.

Since the coin is not fair, (\(P(h)=0.6\) and \(P(t)=0.4\)), then \(P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4\), multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh
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