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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.
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
Answer: Probably that 2 headed coin picked was 7/8.
No idea if this is right, whats the OA?
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.
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.
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?
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