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

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If a certain coin is flipped, the probability that the coin [#permalink]  12 Jan 2008, 03:44
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92% (01:22) correct 7% (01:08) wrong based on 4 sessions
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, 23:34, edited 1 time in total.
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Re: coins [#permalink]  12 Jan 2008, 06:23
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
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Re: coins [#permalink]  12 Jan 2008, 06:29
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.
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Re: coins [#permalink]  12 Jan 2008, 06:44
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???
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Re: coins [#permalink]  12 Jan 2008, 18:57
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.
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Re: coins [#permalink]  12 Jan 2008, 19:18
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
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Re: coins [#permalink]  25 Aug 2008, 09:13
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
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Re: coins [#permalink]  23 Mar 2009, 12:18
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.
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Re: coins [#permalink]  27 Sep 2009, 06:57
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
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Re: coins [#permalink]  29 Sep 2010, 20:38
Berboulli again: 5C5 *(1/2)^5
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Re: coins [#permalink]  01 Aug 2011, 09:28
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.
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Re: coins [#permalink]  01 Aug 2011, 13:47
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..
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Re: coins [#permalink]  20 Aug 2011, 06:17
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?
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  25 Feb 2012, 19:52
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"...)
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  25 Feb 2012, 23:51
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.
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  26 Feb 2012, 05:30
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
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  26 Feb 2012, 05:44
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.
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  26 Feb 2012, 22:25
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
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  26 Feb 2012, 23:53
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fortsill wrote:
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.
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Re: If a certain coin is flipped, the probability that the coin [#permalink]  03 Apr 2012, 21:43
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?
Re: If a certain coin is flipped, the probability that the coin   [#permalink] 03 Apr 2012, 21:43
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