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If a certain coin is flipped, the probability that the coin [#permalink]
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12 Jan 2008, 03:44
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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
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Last edited by Bunuel on 25 Feb 2012, 23:34, edited 1 time in total.
Added the OA



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Re: coins [#permalink]
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12 Jan 2008, 06:23
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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]
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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]
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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^mn apply???



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Re: coins [#permalink]
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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: HH (heads first time, heads second time) 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]
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12 Jan 2008, 19:18
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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



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Re: coins [#permalink]
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25 Aug 2008, 09:13
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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]
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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]
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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]
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29 Sep 2010, 20:38
Berboulli again: 5C5 *(1/2)^5



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If a certain coin is flipped, the probability that the coin will land [#permalink]
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20 Oct 2010, 06:41
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
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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20 Oct 2010, 07:27
(1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/32



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Re: coins [#permalink]
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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]
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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]
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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]
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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]
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25 Feb 2012, 23:51
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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. Answer: E. 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]
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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]
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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]
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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 1p) B = mCn * p^n * q^(mn) ......with m=5, n=3, p=q=1/2




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