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

Question

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

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

Bunuel · Accepted Answer

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

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.

mihir0710 · Answer

Refer to the solution in the picture

srini123 · Answer

Bunuel - the quant maestro --- nails one more problem...

) I missed the multiplying by 5 too

kpali · Answer

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.

Bunuel · Answer

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.

SVaidyaraman · Answer

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.

ScottTargetTestPrep · Answer

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

vanam52923 · Answer

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

Bunuel · Answer

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

Basim2016 · Answer

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

Basshead · Answer

P(heads at least 4 times) = P(4 heads & 1 tails) + P(5 heads)

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

Choice E.

CEdward · Answer

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

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
2. heads 5x : (3/5)^5

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

Answer is E.

Kinshook · Answer

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

sarthak1701 · Answer

2 cases of AT LEAST 4 Heads

HHHHH

and

HHHHT

First case can only be arranged in 1 way so the probability is simply 0.6^5

For the second case there are 5!/4! = 5 permutations (order matters or you would simply be multiplying the probabilities)

Multiply the Permutations by probabilities 0.6^4*0.4*5 (4 heads 1 tails multiplied by possible permutations. Now add this to the probability calculated in the first step.

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

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