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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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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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25 Oct 2009, 18:54
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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

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

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

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25 Oct 2009, 19:12
1
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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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, $$P(h\geq{4})=0.6^5+5*0.6^4*0.4$$.

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

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

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..

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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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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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

Refer to the solution in the picture
Attachments

Solution Head.jpeg [ 48.14 KiB | Viewed 3376 times ]

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

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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.

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

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

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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Posts: 58390
Re: For one toss of a certain coin, the probability that the outcome is he  [#permalink]

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

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

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14 Oct 2019, 04:30
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