Last visit was: 21 May 2024, 02:50 It is currently 21 May 2024, 02:50
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Retired Moderator
Joined: 17 Sep 2013
Posts: 279
Own Kudos [?]: 1225 [76]
Given Kudos: 139
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE:Analyst (Consulting)
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 93365
Own Kudos [?]: 625460 [30]
Given Kudos: 81917
Send PM
avatar
Intern
Intern
Joined: 12 May 2013
Posts: 39
Own Kudos [?]: 101 [18]
Given Kudos: 12
Send PM
General Discussion
User avatar
Manager
Manager
Joined: 02 Mar 2016
Status:Keep it simple stupid!!
Posts: 62
Own Kudos [?]: 146 [2]
Given Kudos: 26
Location: India
Concentration: General Management, Operations
GMAT 1: 680 Q47 V36
GPA: 3.37
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
1
Kudos
1
Bookmarks
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
Current Student
Joined: 30 Dec 2013
Status:preparing
Posts: 28
Own Kudos [?]: 26 [6]
Given Kudos: 26
Location: United Arab Emirates
Concentration: Technology, Entrepreneurship
GMAT 1: 660 Q45 V35
GMAT 2: 640 Q49 V28
GPA: 2.84
WE:General Management (Consumer Products)
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
5
Kudos
1
Bookmarks
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
VP
VP
Joined: 12 Dec 2016
Posts: 1023
Own Kudos [?]: 1789 [0]
Given Kudos: 2562
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
Send PM
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
Intern
Intern
Joined: 19 Dec 2017
Posts: 6
Own Kudos [?]: 2 [1]
Given Kudos: 0
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
1
Kudos
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)
Senior Manager
Senior Manager
Joined: 31 Jul 2017
Posts: 435
Own Kudos [?]: 443 [3]
Given Kudos: 752
Location: Malaysia
GPA: 3.95
WE:Consulting (Energy and Utilities)
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
2
Kudos
1
Bookmarks
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 = \(32\)
\(P = \frac{5C3}{32}\)
Senior Manager
Senior Manager
Joined: 08 Jun 2015
Posts: 259
Own Kudos [?]: 83 [0]
Given Kudos: 145
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33
Send PM
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 !!
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18874
Own Kudos [?]: 22278 [6]
Given Kudos: 285
Location: United States (CA)
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
4
Kudos
2
Bookmarks
Expert Reply
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.

Answer: D
GMAT Club Legend
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21843
Own Kudos [?]: 11688 [4]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
2
Kudos
2
Bookmarks
Expert Reply
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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5350
Own Kudos [?]: 4012 [2]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
1
Kudos
1
Bookmarks
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
VP
VP
Joined: 10 Jul 2019
Posts: 1389
Own Kudos [?]: 549 [0]
Given Kudos: 1656
Send PM
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-
Director
Director
Joined: 09 Jan 2020
Posts: 961
Own Kudos [?]: 226 [0]
Given Kudos: 434
Location: United States
Send PM
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
Intern
Intern
Joined: 21 Sep 2023
Posts: 4
Own Kudos [?]: 0 [0]
Given Kudos: 484
Send PM
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).

Answer: D.

­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?
plz answer
thanks in advance
Math Expert
Joined: 02 Sep 2009
Posts: 93365
Own Kudos [?]: 625460 [0]
Given Kudos: 81917
Send PM
Re: What is the probability of getting exactly three heads on [#permalink]
Expert Reply
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).

Answer: D.

­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?
plz answer
thanks in advance

­
I don't get the difference between a and b. If you toss a fair coin 5 times, the probability of getting 3 heads and 2 tails is 5!/(3!2!)*1/2^5.
GMAT Club Bot
Re: What is the probability of getting exactly three heads on [#permalink]
Moderator:
Math Expert
93364 posts