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

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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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

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

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

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..

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"...)

in this case we can also use the following formula -nCk/2^n = 5C3/2^5=10/32
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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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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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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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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

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]


hey thanks for you answer

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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Yes, that's correct.
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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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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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

Answer: E

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

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


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

Regards,
Fabio.
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