Last visit was: 27 Mar 2025, 22:33 It is currently 27 Mar 2025, 22:33
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
User avatar
marcodonzelli
Joined: 22 Nov 2007
Last visit: 22 Aug 2014
Posts: 627
Own Kudos:
3,017
 [86]
Posts: 627
Kudos: 3,017
 [86]
8
Kudos
Add Kudos
78
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 March 2025
Posts: 100,115
Own Kudos:
Given Kudos: 92,748
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,115
Kudos: 711,434
 [92]
38
Kudos
Add Kudos
53
Bookmarks
Bookmark this Post
avatar
pmenon
Joined: 28 Dec 2005
Last visit: 01 Jul 2009
Posts: 693
Own Kudos:
599
 [25]
Given Kudos: 2
Posts: 693
Kudos: 599
 [25]
15
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
General Discussion
User avatar
hanumayamma
Joined: 30 Jun 2007
Last visit: 14 May 2015
Posts: 366
Own Kudos:
550
 [2]
Posts: 366
Kudos: 550
 [2]
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
The probability of landing heads and not landing on heads is same = ½
The probability of first three heads = ½* ½ * ½
The probability of last two landing not on heads = ½ * ½
The total probability = ½ * ½ * ½ * ½ * ½ = 1/ 32
Answer: E
avatar
srivas
Joined: 27 Oct 2008
Last visit: 28 Sep 2009
Posts: 97
Own Kudos:
Given Kudos: 3
Posts: 97
Kudos: 304
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

Soln: Ans is E: 1/32

Total number of outcomes is = 2 * 2 * 2 * 2 * 2 = 32 ways

Now it says that first 3 heads and 2 tails
{HHHTT} => of the above 32 possible outcomes, in only one outcome will we get {HHHTT}

THus probability is = 1/32
avatar
BenchPrepGURU
Joined: 18 Jul 2011
Last visit: 03 Aug 2011
Posts: 24
Own Kudos:
Given Kudos: 2
Posts: 24
Kudos: 40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bernoulli can't be used here - it will give you the probability of getting 3 heads or 2 tail, not the probability of getting 3 heads followed by two tails. Bernoulli doesn't take order into account, only successes (however you define it) within a certain number of trials. I suppose a misapplication of the Bernoulli formula will yield the correct answer, but it's important to understand how the formula really works if you're going to use it.

Personally, I've never seen a probability question where knowing the Bernoulli formula would give one a significant advantage... I suspect that a lot of time is wasted memorizing such formulas.

The simplest way to think about this is that HHHTT is just one outcome among 32 possible outcomes.
User avatar
gladvijay
Joined: 27 Feb 2011
Last visit: 08 Sep 2011
Posts: 22
Own Kudos:
Given Kudos: 9
Posts: 22
Kudos: 9
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The coin tosses are mutually exclusive.. result of the first coin toss doesnt affect the result of second coin toss..

pr. for heads on first toss = 1/ 2
pr. for heads on second toss = 1/ 2
pr. for heads on third toss = 1/ 2
pr. for tails on 4th toss = 1/ 2
pr. for tails on 5th toss = 1/ 2

= (1/2)^ 5 = 1/ 32

if it was a bag of red balls and blue balls.. then drawing for a red ball would have affected the probability for the next draw.. here is not the case..
User avatar
fortsill
Joined: 24 Feb 2012
Last visit: 07 Jun 2012
Posts: 25
Own Kudos:
Given Kudos: 18
Posts: 25
Kudos: 33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
When I see a coin or dice question, I automatically tend to think using:
- the technique that khan academy references in his tutorial on probabilities (that's the technique that came to my mind first in this case)
- the binomial formula b(n;x,p) - which walker tends to use in a few instances here.

In this case:
5 trials, each trial has 2 possibilities; total outcomes = 2^5 = 32
The possibility we're looking for is a unique outcome: HHHTT - which can only occur in one way.
So, probability = 1/32

Another technique that the khan academy references:
P (H and H and H and T and T) = P(H) * P(H) * P(H) * P(T) * P(T)
= 1/2 * 1/2 * 1/2 * 1/2 * 1/2
= 1/32

I strongly recommend watching the khan academy videos if the sight of coin & dice questions makes you quiver, like it did for me.
(From a GMAT Verbal standpoint, is the above sentence grammatically correct? should it be "make" or "makes" - and i'm struggling with the last part "like it did for me"...)
User avatar
LalaB
User avatar
Current Student
Joined: 23 Oct 2010
Last visit: 17 Jul 2016
Posts: 228
Own Kudos:
Given Kudos: 73
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Posts: 228
Kudos: 1,222
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).

As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.

Hope it's clear.

in this case we can also use the following formula -nCk/2^n = 5C3/2^5=10/32
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 March 2025
Posts: 100,115
Own Kudos:
Given Kudos: 92,748
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,115
Kudos: 711,434
Kudos
Add Kudos
Bookmarks
Bookmark this Post
LalaB
Bunuel
IF THE QUESTION WERE:
A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

Then the answer would be: \(P=\frac{5!}{3!2!}*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}\), since in this case the winning scenario of HHHTT can occur in many ways: HHHTTT, HTHHT, HTTHH, ... which basically is # of permutations of 5 letters out of which 3 H's and 2 T's are identical, so \(\frac{5!}{3!2!}\).

As you can see the probability of the event for modified question is 10 times higher that the probability of the event for the initial question, and that's because # of favorable outcomes is 10 for modified question and 1 for initial question.

Hope it's clear.

in this case we can also use the following formula -nCk/2^n = 5C3/2^5=10/32

Yes, you can: both give exactly the same expression, since \(C^3_5=\frac{5!}{3!2!}\). Though your reasoning for 5C3 could be different from what I used in my approach: selecting which 3 flips out of 5 will give heads up.
User avatar
fortsill
Joined: 24 Feb 2012
Last visit: 07 Jun 2012
Posts: 25
Own Kudos:
Given Kudos: 18
Posts: 25
Kudos: 33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Although for bunuel's extension of the original question, seems like he was using the binomial prob formula...m trials, n successes, prob of success p, prob of failure q ( or 1-p)
B = mCn * p^n * q^(m-n)
......with m=5, n=3, p=q=1/2
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 March 2025
Posts: 100,115
Own Kudos:
711,434
 [3]
Given Kudos: 92,748
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,115
Kudos: 711,434
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fortsill
Although for bunuel's extension of the original question, seems like he was using the binomial prob formula...m trials, n successes, prob of success p, prob of failure q ( or 1-p)
B = mCn * p^n * q^(m-n)
......with m=5, n=3, p=q=1/2

Yes, it's a formula for binomial distribution explained here: math-probability-87244.html

Some questions to practice:
what-is-the-probability-that-a-field-gun-will-hit-thrice-on-127334.html
there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html
combination-ps-55071.html
the-probability-that-a-family-with-6-children-has-exactly-88945.html?

Hope it helps.
avatar
dchow23
Joined: 16 May 2011
Last visit: 18 Feb 2013
Posts: 52
Own Kudos:
Given Kudos: 2
Posts: 52
Kudos: 46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
whats the difference between this question and the following?

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

is it because it could any 3 days between the 4th and 8th?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 March 2025
Posts: 100,115
Own Kudos:
711,434
 [1]
Given Kudos: 92,748
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,115
Kudos: 711,434
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
dchow23
whats the difference between this question and the following?

If the probability of rain on any given day in Chicago during the summer is 50%, independent of what happens on any other day, what is the probability of having exactly 3 rainy days from July 4 through July 8, inclusive?

is it because it could any 3 days between the 4th and 8th?

The difference is explained here: if-a-certain-coin-is-flipped-the-probability-that-the-coin-58357.html#p1049993

The probability of rain each day is 1/2 and the probability of no rain is also 1/2. \(C^3_5=10\) represent ways to choose on which 3 days out of 5 there will be a rain, so \(P=C^3_5*(\frac{1}{2})^3*(\frac{1}{2})^2=\frac{10}{32}=\frac{5}{16}\).

Or think about it this way: we want the probability of the following event: RRRNN, where R represent rain day and N represents no-rain day. Now, each R and each N have individual probability of 1/2, so \((\frac{1}{2})^5\).

But the case of RRRNN can occur in many ways: RRRNN, RRNRRN, RNRRN, NRRRN, ... basically it will be equla to # of arrangements (permutations) of 5 letters RRRNN out of which there are 3 identical R's and 2 identical N's. That # of arrangements is \(\frac{5!}{3!2!}\), (notice that it's the same as \(C^3_5\)). So, finally \(P=\frac{5!}{3!2!}*(\frac{1}{2})^5=\frac{5}{16}\).

For more on this topic check Combinations and Probability chapters of Math Book:
math-combinatorics-87345.html
math-probability-87244.html

Also check similar questions to practice:
if-the-probability-of-rain-on-any-given-day-is-50-what-is-99577.html
on-saturday-morning-malachi-will-begin-a-camping-vacation-100297.html
what-is-the-probability-that-a-field-gun-will-hit-thrice-on-127334.html
there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html
combination-ps-55071.html
the-probability-that-a-family-with-6-children-has-exactly-88945.html?

Hope it helps.
User avatar
Maxirosario2012
Joined: 02 Apr 2012
Last visit: 11 Nov 2016
Posts: 51
Own Kudos:
Given Kudos: 155
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE:Consulting (Consulting)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
My try ( I am a beginner trying to solve this kind of problems using combinations :idea:

\(\frac{C^5_3}{{(C^2_1)^5}}\) = 1/32

\(C^5_3\) : number of combinations in which we obtain 3 H in 5 tosses

\((C^2_1)^5\) : number of combinations of T or H in 5 tosses
avatar
daviddaviddavid
Joined: 26 Mar 2017
Last visit: 20 Jul 2017
Posts: 61
Own Kudos:
Given Kudos: 1
Posts: 61
Kudos: 246
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hey,

could we also calculate it as follows:

(1/2)^5 * (5*4*3)/(3*2*1)[/quote]

Thew correct answer is 1/2^5, so the answer is no. What is the logic behind what you wrote?[/quote]


hey thanks for you answer

I meant to your modified question:
(A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

(1/2)^5 --> probability * (5*4*3)/(3*2*1) --> combinations

many thanks =)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 March 2025
Posts: 100,115
Own Kudos:
Given Kudos: 92,748
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 100,115
Kudos: 711,434
Kudos
Add Kudos
Bookmarks
Bookmark this Post
daviddaviddavid
hey thanks for you answer

I meant to your modified question:
(A certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and not on heads up on 2 flips? (so without fixing order of occurrences of heads and tails)

(1/2)^5 --> probability * (5*4*3)/(3*2*1) --> combinations

many thanks =)

Yes, that's correct.
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,796
Own Kudos:
12,279
 [3]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,796
Kudos: 12,279
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi All,

This prompt is actually a remarkably straight-forward probability question, since it asks you for an EXACT series of results based on coin flips:

HHHTT

Since there's a 1/2 chance of flipping heads and a 1/2 chance of flipping tails, the probability of flipping that exact sequence of events is:

(1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

Final Answer:

GMAT assassins aren't born, they're made,
Rich
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,769
Own Kudos:
33,155
 [2]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,769
Kudos: 33,155
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

We want P(heads on 1st AND heads on 2nd AND heads on 3rd AND tails on 4th AND tails on 5th)
= P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd) x P(tails on 4th) x P(tails on 5th)
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/32

Answer: E

Cheers,
Brent
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 888
Own Kudos:
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 888
Kudos: 1,609
Kudos
Add Kudos
Bookmarks
Bookmark this Post
marcodonzelli
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32
\(? = P\left( {{\text{specific}}\,\,{\text{4}}\,\,{\text{landings}}\,\,{\text{sequence}}} \right) = \frac{1}{{32}}\)

\({\rm{Reason}}:\,\,\,\left\{ \matrix{\\
\,{2^4} = 32\,\,\,{\rm{equiprobable}}\,\,{\rm{possible}}\,\,{\rm{outcomes}} \hfill \cr \\
\,1\,\,\, = \,\,\,{\rm{specific}}\,\,{\rm{4}}\,\,{\rm{landings}}\,\,{\rm{sequence}}\,\,{\rm{desired}}\,\,\,({\rm{favorable)}} \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
 1   2   
Moderators:
Math Expert
100114 posts
PS Forum Moderator
519 posts