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Say you have 5 coins. 1 coin has heads on both sides and the [#permalink]

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Say you have 5 coins. 1 coin has heads on both sides and the other 4 coins are normal (heads on one side, tails on the other)

A coin is selected at random and flipped five times, each time landing on heads. What is the probability that this coin is the coin that has heads on each side?

Say you have 5 coins. 1 coin has heads on both sides and the other 4 coins are normal (heads on one side, tails on the other)

A coin is selected at random and flipped five times, each time landing on heads. What is the probability that this coin is the coin that has heads on each side?

In this probability of \(\frac{9}{40}\) unfair coin contributed \(8\) shares out of \(9\), so the probability that the coin we chose randomly is unfair is: \(\frac{8}{9}\)
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We determined that the probability of five heads is 8/40+1/40=9/40. So, there are 9 chances out of 40 this to happen.

But we know that this probability has already happened, so this 9 chances "worked". For this 9 chances unfair coin contributed 8 shares (8/40 of the probability) and fair coin only 1 (1/40). So the chance that the coin is unfair is 8 out of 8+1=9.

We can calculate this in another way as the probability P(h=5)=9/40, the chances that this is unfair coin would be (1/5)/(9/40)=8/9 and the probability that the coin is fair is (1/40)/(9/40)=1/9.

This problem is dealing with the concept of conditional probability which is advanced issue of probability. I doubt that much of this is needed for GMAT.

Would be nice to see OA and OE for this question.
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There are five coins. Four of them are regular two-sided coins, and one of them has two Heads. (In other words, one of the coins has Heads on both sides.)

You pick a coin at random and flip it five times. It comes up Heads each time.

What is the probability that you picked the coin with two Heads (the coin with Heads on both sides)?

When a regular coin is flipped once it has a 1/2 chance of being heads. So for a regular coin to be heads five times it'll equal (1/2)^5 or 1/32 ... os one out of every 32 tries a coins will come up Heads fives times in a row.

We have 4 regular coins.. therefore 4 chances to hit the 5 heads with the regular coins...therefore 4*(1/32)= 1/8 probability that a regualr coin will be he 5 heads in a row.

Next we setup an equation. We know the probabilty of it hitting 5 heads in a row = 1 (it will happen) and we'll set X to be the probability that we selected the 2 headed coin.

so (1/8) + x = 1

therefore x=7/8

Answer: Probably that 2 headed coin picked was 7/8.

No idea if this is right, whats the OA?
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There are five coins. Four of them are regular two-sided coins, and one of them has two Heads. (In other words, one of the coins has Heads on both sides.)

You pick a coin at random and flip it five times. It comes up Heads each time.

What is the probability that you picked the coin with two Heads (the coin with Heads on both sides)?

Probability of choosing any coin will be 1/5.

Probability of choosing the specific coin with two heads is (just one coin)/(lot of 5 coins) = 1/5.

Probability of choosing a normal coin with one head and one tail ( 4 of the remaining lot of 5 that are good coins)/(lot of 5 coins) = 4/5.

Here's how I look at it: 1/40 is the chance of picking up a non-trick coin and flipping 5H.

Chance of picking up a trick coin is 1/5, or 8/40.

Given that 5H has already happened, what is the probability that the coin is rigged?

Of all the times this event can occur, 8 of the 9 times would be due to a rigged coin. Only 1 of 9 chances would be attributed to pulling this off with a regular coin, if this event occurs.

Probability of choosing the specific coin with two heads is (just one coin)/(lot of 5 coins) = 1/5.

Probability of choosing a normal coin with one head and one tail ( 4 of the remaining lot of 5 that are good coins)/(lot of 5 coins) = 4/5.

That is all - PERIOD!!!

IMO, this would hold if all coins where equal. Given the outcome, 5 consecutive heads, not all the coins are equal.

they all have an equal chance of being selected but they DO NOT all have an equal chance of satisfying the 5H criteria.

The question is actually confusing.

It says pick a coin first. Then flip it five times and you get Heads everytime.

If you choose the bad coin, you are bound to get head everytime. So probability of getting a head with the bad coin is always 1 and probability of getting a tails is 0.

All you have got to do is pick the bad coin and the chances are 1/5. As you pick it, you will get five heads no matter what.

If the question was rephrased say, all coins are good with equal probability for a heads as 1/2 and tails as 1/2 as well, then the situation would change. You'd have to choose any one coin in 1/5 ways and then flip it 5 times to get 5 heads, for this you'd have to use the Bernoulli's trials concept. For the above situation as well, bernoulli's trials is applicable however, since probablity of getting heads is always 1, it ends up at 1/5.

Anyone with a better logic, I'd certainly appreciate it.
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I dunno BarneyStinson, the way the question was worded seemed quite clear to me.

I also initially had trouble with it and (unfortunately) it took me more than 2 minutes to come up with the answer.

I drew a small probability tree as a visual guide. Obviously I didn't draw each flip, but only the whether the "desired" outcome us achieved (5H in a row).

The first coin has a probability of 1 of getting 5H if it is chosen, while it is 1/32 for the others. Next, I multiplied each probability by 1/5 (the chance of randomly choosing that coin). Finally, I divided 1/5 (1 multiplied by 1/5 - the "desired outcome") by (1/5 + 4/(32*5) - the total amount of outcomes) to get 8/9.

Conditional probability is a tricky concept, one thing good about it is that it's not tested in GMAT.

General case:

There are \(k\) coins: \(f\) are fair coins and \(c\) are counterfeit, two-headed coins (\(f+c=k\)). One coin is chosen at random and tossed \(n\) times. The results are all heads. What is the probability that the coin tossed is the two-headed one?

The probability that the coin tossed is two-headed: \(\frac{\frac{c}{k}*1^n}{\frac{f+c2^n}{k2^{n}}}=\frac{c2^n}{f+c2^n}\).

Now, if you substitute \(c\), \(f\) and \(n\) by the values from the original question you'll get the probability of \(\frac{8}{9}\).
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