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

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12 Jan 2008, 02:44

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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?

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.

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.

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

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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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.

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}

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

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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
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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.

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..

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

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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"...)

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.

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

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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
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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, 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

gmatclubot

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

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