If a certain coin is flipped, the probability that the coin : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 19 Feb 2017, 14:26

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If a certain coin is flipped, the probability that the coin

Author Message
TAGS:

### Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 501 [0], given: 0

If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

12 Jan 2008, 02:44
7
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

85% (01:28) correct 15% (00:41) wrong based on 678 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Feb 2012, 22:34, edited 1 time in total.
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 150 [2] , given: 2

### Show Tags

12 Jan 2008, 05:23
2
KUDOS
2
This post was
BOOKMARKED
Its like this:

On the first three flips, you must get heads. Whats the probability of getting heads ? Its 1/2

So for the first three flips, your probability is (1/2)^3 = 1/8

Now for the last two, you want to get tails only. Whats the prob of getting tails ? Well, its the same as prob of getting a heads, namely, 1/2

For the last two flips, your probability is (1/2)^2 = 1/4

So your overall probability for the event in question is 1/8*1/4 = 1/32
Manager
Joined: 01 Sep 2007
Posts: 100
Location: Astana
Followers: 1

Kudos [?]: 26 [0], given: 0

### Show Tags

12 Jan 2008, 05:29
yeah i was applying the same logic in the end.
but when I doubt, i think this what is the probability that exactly three heads and exactly 2 tails will show up? probably not much, and definetely less that 1/2.
VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 8

Kudos [?]: 501 [0], given: 0

### Show Tags

12 Jan 2008, 05:44
pmenon wrote:
Its like this:

On the first three flips, you must get heads. Whats the probability of getting heads ? Its 1/2

So for the first three flips, your probability is (1/2)^3 = 1/8

Now for the last two, you want to get tails only. Whats the prob of getting tails ? Well, its the same as prob of getting a heads, namely, 1/2

For the last two flips, your probability is (1/2)^2 = 1/4

So your overall probability for the event in question is 1/8*1/4 = 1/32

it's easier than what I tought..I got rid of this choice because it seems too easy. why doesn't the formula N=nCm*p^n * q^m-n apply???
Intern
Joined: 29 Jun 2007
Posts: 49
Followers: 0

Kudos [?]: 9 [0], given: 0

### Show Tags

12 Jan 2008, 17:57
Hmmm. I choose E too.

But basically I have:

P=number of favorable outcomes/total number of possible outcomes.

So my favorable outcome is 1 that is HHHTT,
and possible outcome is 2^5.

Think of it this way, If you only flip the coin twice, you have four possible outcomes:
TT (tails first time, tails second time)
HT (heads first time, tails second time)
TH (tails first time, heads first time)
Or 2^2

If you flip three time, 2^3...four time, 2^4, etc.
Director
Joined: 30 Jun 2007
Posts: 792
Followers: 1

Kudos [?]: 157 [0], given: 0

### Show Tags

12 Jan 2008, 18:18
1
This post was
BOOKMARKED
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
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 35

Kudos [?]: 877 [0], given: 5

### Show Tags

25 Aug 2008, 08:13
1
This post was
BOOKMARKED
marcodonzelli wrote:
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

Probability of HHHTT = (1/2 *1/2 *1/2 )*(1/2 *1/2 )
=1/32
_________________

Smiling wins more friends than frowning

Senior Manager
Joined: 06 Jul 2007
Posts: 285
Followers: 3

Kudos [?]: 51 [0], given: 0

### Show Tags

23 Mar 2009, 11:18
x2suresh wrote:
marcodonzelli wrote:
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

Probability of HHHTT = (1/2 *1/2 *1/2 )*(1/2 *1/2 )
=1/32

There are 32 possible ways and only one favourable way (HHHTT). The probability is 1/32.
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 145 [0], given: 3

### Show Tags

27 Sep 2009, 05:57
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
Intern
Joined: 07 Sep 2010
Posts: 2
Location: Chicago, IL
Schools: Booth, Kellog
WE 1: 10 yrs - Technology and Finance
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

29 Sep 2010, 19:38
Berboulli again: 5C5 *(1/2)^5
Manager
Joined: 06 Feb 2010
Posts: 176
Schools: University of Dhaka - Class of 2010
GMAT 1: Q0 V0
GPA: 3.63
Followers: 62

Kudos [?]: 1192 [0], given: 182

If a certain coin is flipped, the probability that the coin will land [#permalink]

### Show Tags

20 Oct 2010, 05:41
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
_________________

Practice Makes a Man Perfect. Practice. Practice. Practice......Perfectly

Critical Reasoning: http://gmatclub.com/forum/best-critical-reasoning-shortcuts-notes-tips-91280.html

Collections of MGMAT CAT: http://gmatclub.com/forum/collections-of-mgmat-cat-math-152750.html

MGMAT SC SUMMARY: http://gmatclub.com/forum/mgmat-sc-summary-of-fourth-edition-152753.html

Sentence Correction: http://gmatclub.com/forum/sentence-correction-strategies-and-notes-91218.html

Arithmatic & Algebra: http://gmatclub.com/forum/arithmatic-algebra-93678.html

I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Intern
Joined: 18 Sep 2010
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: If a certain coin is flipped, the probability that the coin will land [#permalink]

### Show Tags

20 Oct 2010, 06:27
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/32
Intern
Joined: 18 Jul 2011
Posts: 49
Followers: 3

Kudos [?]: 18 [0], given: 2

### Show Tags

01 Aug 2011, 08:28
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.
Intern
Joined: 27 Feb 2011
Posts: 48
Followers: 0

Kudos [?]: 3 [0], given: 9

### Show Tags

01 Aug 2011, 12:47
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..
Intern
Joined: 17 Aug 2011
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

20 Aug 2011, 05:17
1/2^5 X 5C3/(5!/3!2!) = 1/32

generic probability of 3h and 2t over the possible permutations were hhh represent the same outcome ad tt represent the same outcome (5!/2!3!)

make sense?
Intern
Joined: 24 Feb 2012
Posts: 33
Followers: 0

Kudos [?]: 15 [0], given: 18

Re: If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

25 Feb 2012, 18:52
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"...)
Math Expert
Joined: 02 Sep 2009
Posts: 37024
Followers: 7226

Kudos [?]: 96053 [3] , given: 10706

Re: If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

25 Feb 2012, 22:51
3
KUDOS
Expert's post
7
This post was
BOOKMARKED
marcodonzelli wrote:
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 the probability that coin will land heads up on the first 3 flips and not on the last 2 flips. So there is ONLY one favorable outcome, namely heads up on the first 3 flips and tails up on the last 2 flips: HHHTT.

# of total out comes is 2^5=32.

P=favorable/total=1/32.

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.
_________________
Senior Manager
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 22

Kudos [?]: 338 [0], given: 73

Re: If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

26 Feb 2012, 04:30
Bunuel wrote:
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
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Math Expert
Joined: 02 Sep 2009
Posts: 37024
Followers: 7226

Kudos [?]: 96053 [0], given: 10706

Re: If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

26 Feb 2012, 04:44
LalaB wrote:
Bunuel wrote:
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.
_________________
Intern
Joined: 24 Feb 2012
Posts: 33
Followers: 0

Kudos [?]: 15 [0], given: 18

Re: If a certain coin is flipped, the probability that the coin [#permalink]

### Show Tags

26 Feb 2012, 21:25
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
Re: If a certain coin is flipped, the probability that the coin   [#permalink] 26 Feb 2012, 21:25

Go to page    1   2    Next  [ 28 posts ]

Similar topics Replies Last post
Similar
Topics:
What is the probability of flipping a fair coin two times and the coin 1 13 Dec 2016, 06:18
2 If a certain coin is flipped, each of the probabilities that the coin 3 26 Sep 2016, 04:39
6 Jason flips a coin three times. What is the probability that the coin 4 13 Aug 2015, 09:35
4 If six coins are flipped simultaneously, the probability of 7 23 Jan 2014, 12:37
6 When a certain coin is flipped, the probability of heads is 4 28 Oct 2012, 17:47
Display posts from previous: Sort by