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

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Re: For one toss of a certain coin, the probability that the outcome is he [#permalink]
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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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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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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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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 outcome is he [#permalink]
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Given: For one toss of a certain coin, the probability that the outcome is heads is 0.6.
Asked: If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

The probability that the outcome will be heads at least 4 times = The probability that the outcome will be heads 4 times + The probability that the outcome will be heads 5 times = 5C4*(.6)^4*(.4) + 5C5*(.6)^5 = 5(.6)^4(.4) + (.6)^5

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

DEAR FELLOW,

ORDER depends on the purpose.
1. Same purpose: order does NOT matter (0.6)^5......same selected, same purpose.

2. Different Purpose...... Order matters ...... 5!/4!1! X (0.6)^4 x (0.4) .... 2 different selections, so purpose is different.

It is very late to reply may be Mr. Kpali has taken gmat with distinction but it is never late for learners like me.

Best Regards
Basim

Posted from my mobile device

Originally posted by Basim2016 on 19 Aug 2020, 23:56.
Last edited by Basim2016 on 20 Aug 2020, 00:06, edited 2 times in total.
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Re: For one toss of a certain coin, the probability that the outcome is he [#permalink]

P(4 heads & 1 tails) = (0.6)^4 * 0.4 * 5C4 = 5 (0.6)^4 * 0.4
P(heads at least 4 times) = 5 (0.6)^4 * 0.4 * 0.6^5

Choice E.
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Re: For one toss of a certain coin, the probability that the outcome is he [#permalink]
This took me unnecessarily

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

"at least 4 times" means one of the following two scenarios:

1. heads 4x + tails 1x: (3/5)^4 x 2/5 x 5!/4! = (0.6)^4 x 2

(0.6)^4 x 2 + (3/5)^5 = 5(0.6)^4(0.4)+(0.6)^5

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