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Probability Problems (Confused!)

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Probability Problems (Confused!)

by thp510 » Sun Oct 17, 2010 9:44 am
I'm having problems with "Probability Strategy". Can someone help explain why the answers are what they are?

Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"
Book Answer: Draw it out. = 3/8.
How did they get this number/answer? I originally thought (1/2)(1/2)(1/2)= 1/8 chances. However, this is wrong because it's only asking for EXACTLY twice. So what's the math equation instead of just drawing it all out and spending too much time?

Question 2:
"What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?"
Book Answer: Use the 1-x method (probability of success + probability of failure = 1). The probability of failure or the chance that the dice WON'T roll a six is 5/6. So we're rolling three times, therefore: (5/6)(5/6)(5/6) = 125/216. Now subtract that from 1 and you have 1-(125/126) = 91/216
Question: If this is the case, why can't I just go with multiplying three times the chance that the roll WILL BE a six, so it's (1/6)(1/6)(1/6) = 1/216. This isn't correct. Why?

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"
Book Answer: =6/30. Draw out a probability tree. I understood how they got it this way, but this takes WAY TOO MUCH time.
Question: What's the computation look like? I thought I would take 4 over 6 (sweet candies) and multiply it to 1 over 2 (sour candy) to equal 4 over 12 or (1/3)


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by shovan85 » Sun Oct 17, 2010 10:10 am
thp510 wrote: Question 2:
"What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?"
Book Answer: Use the 1-x method (probability of success + probability of failure = 1). The probability of failure or the chance that the dice WON'T roll a six is 5/6. So we're rolling three times, therefore: (5/6)(5/6)(5/6) = 125/216. Now subtract that from 1 and you have 1-(125/126) = 91/216
Question: If this is the case, why can't I just go with multiplying three times the chance that the roll WILL BE a six, so it's (1/6)(1/6)(1/6) = 1/216. This isn't correct. Why?
If you calculate 1/216 that implies all the three rolls MUST have 6 as a result. Or we can say u will get a definite (6,6,6).
But here question asks at least one of them will be 6. So the cases are
(x,x,6), (x,6,x), (6,x,x) and as it is mentioned AT LEAST this x can be anything from 1,2,3,4,5,6.

If u wanna try your way then it will be very much clumsy. Still let me show you how it works then its upto you if u wanna follow 1-x or not ;)

Take (x,x,6) possibility:
(two failures AND one success) OR (one failure AND two success ) OR (Three Success) [Failure not 6, Success 6]
= 5/6 * 5/6 * 1/6 + 5/6 * 1/6 * 1/6 + 1/6 * 1/6 * 1/6 = 31/216 ...... (1)

Take (x,6,x) possibility: (Here the [6,6,6] possibilty we will discard as it has been taken care in (1))
(failure AND success AND failure) OR (failure AND success AND success) [Discard Three scucesses here as it has been taken care of]
= 5/6 * 1/6 * 5/6 + 5/6 * 1/6 * 1/6 = 30/216 .... (2)

Take (6,x,x) possibilty again discarding (6,6,6) = 30/216 ..... (3)

So answer will be 31/216 + 30/216 + 30/216 = 91/216

So, when asked AT LEAST then calculate the number of FAILURES and SUBTRACT it from 1 .

Hope this helps.
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by shovan85 » Sun Oct 17, 2010 10:28 am
thp510 wrote:
Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"
Book Answer: Draw it out. = 3/8.
How did they get this number/answer? I originally thought (1/2)(1/2)(1/2)= 1/8 chances. However, this is wrong because it's only asking for EXACTLY twice. So what's the math equation instead of just drawing it all out and spending too much time?
I would always prefer the Drawing method as it gives you a crystal clear idea of what you are dealing with. As coins have only two possibilities and if it is 3 or 4 coins there is no harm in following so.

However, if you want to solve it in equation way, still you need a skeletal view of the same.

Question is asking exact two Heads.
It can be (H,H,T) or (H,T,H) or (T,H,H)
= [P(H)*P(H)*P(T)] + [P(H) * P(T) * P(H)] + [P(T) * P(H) * P(H)]
= [1/2 * 1/2 * 1/2] + [1/2 * 1/2 * 1/2] + [1/2 * 1/2 * 1/2]
= 1/8 + 1/8 + 1/8
= 3/8
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by shovan85 » Sun Oct 17, 2010 10:39 am
thp510 wrote: Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"
Book Answer: =6/30. Draw out a probability tree. I understood how they got it this way, but this takes WAY TOO MUCH time.
Question: What's the computation look like? I thought I would take 4 over 6 (sweet candies) and multiply it to 1 over 2 (sour candy) to equal 4 over 12 or (1/3)
Exactly 1 has to be Sour from the pick.

So the possibilities are (Sweet,Sour) or (Sour,Sweet) [These two cases Jack will have EXACTLY 1 sour candy and also This sequence of "Sweet and Sour" or "Sour and Sweet" matters as for the first pick total possibility is 6 and for the second the total possibilty is 5 as from the first pick one candy has been in Jack's Hand]
= [P(Sw)*P(So)] + [P(So)*P(Sw)]
= (4/6 * 2/5) + (2/6 * 4/5)
= 16/30

Please clarify the solution as you have mentioned it as 6/30.
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by thp510 » Sun Oct 17, 2010 12:58 pm
You're awesome! Thanks for the replies.
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by shovan85 » Sun Oct 17, 2010 1:03 pm
thp510 wrote:You're awesome! Thanks for the replies.
My pleasure :)
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by GMATGuruNY » Sun Oct 17, 2010 3:28 pm
Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"

P(HHT) = 1/2 * 1/2 * 1/2 = 1/8.
Since T can occur on the 1st, 2nd or 3rd flip, we multiply by 3:
3*(1/8) = 3/8.

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"

P(sweet,sour) = (4/6)*(2/5) = 4/15.
Since (sour,sweet) also is a good outcome, we multiply by 2:
2*(4/15) = 8/15.

Hope this helps!
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by thp510 » Sun Oct 17, 2010 8:45 pm
GMATGuruNY wrote:Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"

P(HHT) = 1/2 * 1/2 * 1/2 = 1/8.
Since T can occur on the 1st, 2nd or 3rd flip, we multiply by 3:
3*(1/8) = 3/8.

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"

P(sweet,sour) = (4/6)*(2/5) = 4/15.
Since (sour,sweet) also is a good outcome, we multiply by 2:
2*(4/15) = 8/15.

Hope this helps!

Mitch thanks! Your approach is very short and fast. Is this based on some universal equation that wasn't hit upon or just your ability to know math like the back of your hand? I'm amazed.
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