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A bag contains 3 coins, two of them are FAIR and the third [#permalink]
01 Aug 2003, 13:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A bag contains 3 coins, two of them are FAIR and the third one is a DOUBLE HEADED. A coin is selected from the bag at random and then tossed. If head appears the same coin is tossed again; if tail appears another coin is selected at random from the two reamining coins in the bag. What is the probability that
(1) a head appears on the first toss ?
(2) a head appears on both tosses ?
(3) a tail appears on both tosses ? _________________
Re: Probability with a twist [#permalink]
01 Aug 2003, 18:33
Brainless wrote:
A bag contains 3 coins, two of them are FAIR and the third one is a DOUBLE HEADED. A coin is selected from the bag at random and then tossed. If head appears the same coin is tossed again; if tail appears another coin is selected at random from the two reamining coins in the bag. What is the probability that
(1) a head appears on the first toss ? (2) a head appears on both tosses ? (3) a tail appears on both tosses ?
This is pretty simple if you do a probability tree. Especially since both (2) and (3) are conditional on the results of (1). _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
I did not understand Akamaibrah's point of 3rd condition being dependent on the first. That is one way to look at it (1-P(head on first toss) = P (tail on first toss)), however it is not dependent on the first condition.
I did not understand Akamaibrah's point of 3rd condition being dependent on the first. That is one way to look at it (1-P(head on first toss) = P (tail on first toss)), however it is not dependent on the first condition.
You are saying the (2) has the same probability as (1)? Hmmm. That means the second head is guaranteed to come up once the first head does. Is that what you are trying to say?
If a head appears on the first toss, then it is impossible to get two tails. Said in another way, you cannot get two tails unless you first get one tail. Hence, the event in (3) is most certainly dependent on (1) in exactly the same way (2) is. _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
(a) For condition 2 , the second head is NOT sure if the first toss is head. If the first toss is head, and all three coins were fair - the P(second toss=head) will be 1/2. However we have one coin unfair and therefore our answer will be > 1/2. So Eliminate choices in GMAT if this question comes up.
Now for solution of condition 2. IF the first choice is head then
a1 --> Either the coin chosen in first draw is fair - 2/3 chance or
a2 --> Coin chosen in first draw is unfair - 1/3 chance
for a1 -> heads will come up 1/2 times
for a2 -> heads will come up always
Net Prob = 2/3 *1/2 + 1/3 =2/3.
(b) condition 3 is - I am not sure with your argument - not dependent on condition 1. A tail can happen if a fair coin is chosen in the first draw. The chance is 2/3 (2 coins fair). Now the chance of getting a tail on first toss is 1/2 for this fair coin chosen. Hence P (tail in first toss) =2/3*1/2 =1/3.
Now another coin is chosen. Chance that fair coin is cosen =1/2. Chance that tail appears on the second toss with newly chosen fair coin =1/2. therefore P(tail in second toss) =1/4.
Net prob = 1/3*1/4 =1/12
(a) For condition 2 , the second head is NOT sure if the first toss is head. If the first toss is head, and all three coins were fair - the P(second toss=head) will be 1/2. However we have one coin unfair and therefore our answer will be > 1/2. So Eliminate choices in GMAT if this question comes up. Now for solution of condition 2. IF the first choice is head then a1 --> Either the coin chosen in first draw is fair - 2/3 chance or a2 --> Coin chosen in first draw is unfair - 1/3 chance
for a1 -> heads will come up 1/2 times for a2 -> heads will come up always
Net Prob = 2/3 *1/2 + 1/3 =2/3.
(b) condition 3 is - I am not sure with your argument - not dependent on condition 1. A tail can happen if a fair coin is chosen in the first draw. The chance is 2/3 (2 coins fair). Now the chance of getting a tail on first toss is 1/2 for this fair coin chosen. Hence P (tail in first toss) =2/3*1/2 =1/3.
Now another coin is chosen. Chance that fair coin is cosen =1/2. Chance that tail appears on the second toss with newly chosen fair coin =1/2. therefore P(tail in second toss) =1/4. Net prob = 1/3*1/4 =1/12
For number 1:
There are two paths to get a head on the first try.
The first is to pick the two-handed coin (1/3) of the time in which case you are guaranteed to pick head.
The second is to pick one of the two-headed coins (2/3) for either of which the chance of getting heads is (1/2) so 2/3 * 1/2 = 1/3
Add the two together and we get 1/3 + 1/3 = 2/3.
Or we can say: For the six coins, 4 of the faces have heads and 2 have tails. Since any of the 6 faces are equally likely to be shown, the probability of getting heads is 4/6 or 2/3.
For number 2
First recall that the probability is 1/3 that the 2=headed coin is flipped first. If so, then for sure it is heads. Now we pick from 2 fair coins for the second flip. There are 4 sides, 2H and 2T equally likely to come up so the chances of getting heads on the second toss are 1/2. Hence, the chance of the 2-sided coin coming up first and the second coin being heads are 1/3 * 1/2 = 1/6
The probability is 2/3 that the first coin flipped is fair, and if so, the probability of it being a head is 1/2. Hence, the probability of the first coin being fair and a head is 2/3 * 1/2 = 1/3.
If the first coin is fair, then one of the two remaining coins is two headed, hence, there are 3H and 1T available, each equally likely. So the probability of the 2nd coin being heads is 3/4. Hence the probability of the first coin being fair and heads, and the second coin being heads is 1/3 * 3/4 = 1/4.
Finally, the probability of both coins being heads is 1/6 + 1/4 = 5/12.
3). The probability of the first coin being tails is: The prob that it is fair x the probability that a tail will come up or 2/3 * 1/2 = 1/3. Since we needed a fair coin to get a tail on the first flip, one of the remaining coins must be the two-headed one. Hence, we have 3H and 1T available for the second flip, or 1/4 probabiltiy taht it is tails.
Hence the probability that the two coins are tails is: 1/3 * 1/4 = 1/12. _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
I guess there has been a smal mistake while interpreting the question when U got the answere 5/12 for the second case ( that both tossed outcomes are heads )
It says that " INcase there is head in the first TOSS then u RETAIN the same coin and toss again and do not select"
in that case
the probability of having two heads is :
probability that fair coin is selected and both tosses result in heads..
2/3 * 1/4
probability that unfair coin is slected and two heads is 1/3
so we have 1/2 as probability of having HH as the outcome..
Your case would be true if u were to change coins after every toss..... _________________
I guess there has been a smal mistake while interpreting the question when U got the answere 5/12 for the second case ( that both tossed outcomes are heads )
It says that " INcase there is head in the first TOSS then u RETAIN the same coin and toss again and do not select"
in that case the probability of having two heads is : probability that fair coin is selected and both tosses result in heads..
2/3 * 1/4
probability that unfair coin is slected and two heads is 1/3
so we have 1/2 as probability of having HH as the outcome..
Your case would be true if u were to change coins after every toss.....
Thanks.
You are correct that I misread the question.
I concur that the probability of getting 2 heads under the proper rules is indeed 1/2. Nice catch. _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
I guess there has been a smal mistake while interpreting the question when U got the answere 5/12 for the second case ( that both tossed outcomes are heads )
It says that " INcase there is head in the first TOSS then u RETAIN the same coin and toss again and do not select"
in that case the probability of having two heads is : probability that fair coin is selected and both tosses result in heads..
2/3 * 1/4
probability that unfair coin is slected and two heads is 1/3
so we have 1/2 as probability of having HH as the outcome..
Your case would be true if u were to change coins after every toss.....
Thanks.
You are correct that I misread the question. The rules as given actually simplify the problem.
I concur that the probability of getting 2 heads under the proper rules is indeed 1/2. Nice catch. _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993
I tried all means , but I always got 1/2 for (2) and 1/12 for (3) ..
I was busy with my business tours and could not post the answers intme
Did the "official" give explanations. I sure would like to see them.... _________________
Best,
AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993