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

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

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26 Feb 2012, 22:53
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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]

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03 Apr 2012, 20:43
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?
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Re: If a certain coin is flipped, the probability that the coin [#permalink]

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03 Apr 2012, 23:39
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 [#permalink]

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10 Jul 2013, 08:54
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 [#permalink]

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06 Mar 2016, 09:30
Bunuel 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

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.

In this problem is the above the equivalent of some of these other answers (and how I approached the problem) of simply doing

Chance for tails & chance for heads both = 1/2

Chance of heads 1/2 * chance of heads *chance of heads 1/2 * chance of heads 1/2 * chance of tails 1/2 * chance of tails 1/2?

I see the above total as 2^5 over 1 possibility so the numbers are the same but for straight forward problems such as this one is the listed 1/2*1/2*1/2 an acceptable approach or just a coincidence it works for this problem?
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Re: If a certain coin is flipped, the probability that the coin will land [#permalink]

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06 Jan 2017, 22:07
This question is simple since the coin is ideal(i.e., 0.5 probability for either event)
The question asks us to find the probability of the occurrence of the event HHHTT
This is just the product,
P(H)*P(H)*P(H)*P(T)*P(T)=(0.5)*(0.5)*(0.5)*(0.5)*(0.5)
=1/32
If the coin was skewed,i.e., say probability of landing head is 0.75,then the answer would have been,
P(H)*P(H)*P(H)*P(T)*P(T)=(0.75)*(0.75)*(0.75)*(0.25)*(0.25)
=27/1024
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Re: If a certain coin is flipped, the probability that the coin will land   [#permalink] 06 Jan 2017, 22:07

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