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

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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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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
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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: If a certain coin is flipped, the probability that the coin will land [#permalink]
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: If a certain coin is flipped, the probability that the coin will land [#permalink]
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: If a certain coin is flipped, the probability that the coin will land [#permalink]
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 will land [#permalink]
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 will land [#permalink]
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 will land [#permalink]
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 will land [#permalink]
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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 will land [#permalink]
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?
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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dchow23 wrote:
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.
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
My try ( I am a beginner trying to solve this kind of problems using combinations

$$\frac{C^5_3}{{(C^2_1)^5}}$$ = 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
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]
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]

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

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

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

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

$$? = P\left( {{\text{specific}}\,\,{\text{4}}\,\,{\text{landings}}\,\,{\text{sequence}}} \right) = \frac{1}{{32}}$$

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

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

Regards,
Fabio.
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