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# What is the probability of getting exactly three heads on

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Re: What is the probability of getting exactly three heads on [#permalink]
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10

Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32

Total probability = 1/32 x 10 = 5/16
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Re: What is the probability of getting exactly three heads on [#permalink]
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Total number of ways to get 3 heads and 2 tails ( p(h) =p(t)= 1/2 for each)
=2^5=32

# of ways 3 heads can be arranged in 5 tosses= 5c3=10
10/32=5/16 ANS
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Re: What is the probability of getting exactly three heads on [#permalink]
2 ways:
1/ 5C2 / 2^5 => there are totally 2^5 different results, but only 5C2 favor results,
1/ (1/8) * (1/4) * 5C2
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Re: What is the probability of getting exactly three heads on [#permalink]
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HHHTT in any sequence would be required
Probability would be 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 5C3 = 1/32 x 10 = 5/16 (03 heads could be in any position, so 5C3)
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Re: What is the probability of getting exactly three heads on [#permalink]
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

No. of ways of selecting 3 Heads = $$5C3$$
Total number = $$32$$
$$P = \frac{5C3}{32}$$
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Re: What is the probability of getting exactly three heads on [#permalink]
The answer is (5c2)*((1/2)^3)*((1/2)^2). Option D or 5/16 it is !!
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Re: What is the probability of getting exactly three heads on [#permalink]
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

When a fair coin is flipped 5 times, there are 2^5 = 32 possible outcomes. Thus, each possible outcome is equally likely, with probability of 1/32.

The number of possible outcomes for getting 3 heads and 2 tails is 5!/(3! x 2!) = (5 x 4)/2 = 10.

Thus, the probability of getting 3 heads and 2 tails in 5 flips is (1/32) x 10 = 10/32 = 5/16.

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Re: What is the probability of getting exactly three heads on [#permalink]
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Hi All,

We're asked for the probability of getting EXACTLY three heads on five flips of a fair coin. This question can be approached in a couple of different ways, but they all involve a bit of 'Probability math.'

To start, since each coin has two possible outcomes, there are (2)(2)(2)(2)(2) = 32 possible outcomes from flipping 5 coins. To find the number of outcomes that are EXACTLY 3 heads, you can either use the Combination Formula or do some 'brute force' math and map out all of the possibilities.

By choosing 3 heads from 5 tosses, we can use the Combination Formula: N!/(K!)(N-K)! = 5!/(3!)(5-3)! = (5)(4)(3)(2)(1)/(3)(2)(1)(2)(1) = (5)(4)/(2)(1) = 10 possible ways to flip 3 heads from 5 tosses.

You could also list out the options:
HHHTT
HHTHT
HTHHT
THHHT

HHTTH
HTHTH
THHTH

HTTHH
THTHH

TTHHH

Either way, you have 10 total options that fit what we're looking for out of a total of 32 outcomes. 10/32 = 5/16

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Re: What is the probability of getting exactly three heads on [#permalink]
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JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

Favourable cases = 5C3 = 10
Total cases = 2^5 = 32

Probability = 10/32 = 5/16

IMO D

Posted from my mobile device
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Re: What is the probability of getting exactly three heads on [#permalink]
1st)

You need to realize that the Outcome of getting 3 Heads and 2 Tails can happen in various ways, each requiring a Separate Probability that we need to Add as an "OR" Probability

Successful Outcomes - you can have:

H - H - H - T - T
H - H - T - H - T
H - H - T - T - H
etc.

The Number of Ways to Arrange 5 Elements, in which 3 are Indistinguishable H's and 2 are Indistinguishable T's =

5! / (3! * 2!) = 10 Arrangements

AND

2nd)
The Probability of Getting a Heads = P(H) = 1/2
The Probability of Getting a Tails = P(T) = 1/2

Thus, for ANY 1 of the 10 Ways that we can get a Successful Outcome of EXACTLY 3 Heads, the Probability will be the Same:

(1/2) * (1/2) * (1/2) * (1/2) * (1/2) = (1/2)^5

Finally:

Probability of getting EXACTLY 3 heads out of 5 Coin Flips =

(10) * (1/2)^5 = 10 / 32 = 5 / 16

-D-
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Re: What is the probability of getting exactly three heads on [#permalink]
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/32

This means there are 32 possible outcomes.

We are looking for exactly three heads; thus, we the number of ways we can choose 3 heads from 5 coin flips is 5C3= 10 favorable outcomes

Answer = 10 favorable outcomes / 32 total outcomes: 10/32 = 5/16
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Re: What is the probability of getting exactly three heads on [#permalink]
Bunuel wrote:
JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

$$P(HHHTT)=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}$$, we need to multiply by $$\frac{5!}{3!2!}$$ because HHHTT outcome can occur in several ways: HHHTTT, HHTHT, HTHHT, ..., TTHHH ($$\frac{5!}{3!2!}$$ is permutation of 5 letters HHHTT).

­Bunuel
according to this method for numerator we do 5!/3!*2! and for denominator we do (1/2) raised to 3 and (1/2) raised to 3
so my question is, whether following both conditions and their answers and specially their method for solvings are correct or not
1. if a coin is tossed 5 times
a. how many ways 3H and 2T can be selected ...answer 5C2 / 2 raised to 5 that is 10/32
b. how many wasys 3H and 2T can be arranged ....answer...5!/3!*2! divided by 32 that is 10/32
is my approach correct?
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Re: What is the probability of getting exactly three heads on [#permalink]
athudi1507 wrote:
Bunuel wrote:
JusTLucK04 wrote:
What is the probability of getting exactly three heads on five flips of a fair coin?

(A) 1/32
(B) 3/32
(C) 1/4
(D) 5/16
(E) 1/2

I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option

$$P(HHHTT)=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}$$, we need to multiply by $$\frac{5!}{3!2!}$$ because HHHTT outcome can occur in several ways: HHHTTT, HHTHT, HTHHT, ..., TTHHH ($$\frac{5!}{3!2!}$$ is permutation of 5 letters HHHTT).

­Bunuel
according to this method for numerator we do 5!/3!*2! and for denominator we do (1/2) raised to 3 and (1/2) raised to 3
so my question is, whether following both conditions and their answers and specially their method for solvings are correct or not
1. if a coin is tossed 5 times
a. how many ways 3H and 2T can be selected ...answer 5C2 / 2 raised to 5 that is 10/32
b. how many wasys 3H and 2T can be arranged ....answer...5!/3!*2! divided by 32 that is 10/32
is my approach correct?