jsnipes wrote:I thought I could use the slot method on all 'at least one' problems but it seems to be failing me here (or very possibly I am doing it wrong).
Herb bought a different present for each of his four grandchildren, but he forgot to label the gifts after wrapping them. If Herb randomly gives each grandchild a present, what is the probability that at least one grandchild gets the correct gift?
23/24
7/8
5/8
4/8
3/8
So, seeing "at least one" I decide to use the "1-P" as well as the slot method. So, I tried to find the probability that no grandchildren get the correct present and attempted to do this using the slot method.
The first GC in slot one will get the present that is not his 3/4 times since there are 4 presents and 3 are not for him. GC 2 will now get a present that is not his 2 out of the remaining 3 presents left, or 2/3. GC 3 will then get the wrong present approximately 1/2 the time as will the 4th GC.
So, 3/4 * 2/3 * 1/2 * 1/2=6/48 or 1/8. 1- 1/8=7/8 or answer B however the answer in the book says the correct answer is answer C but only explains that it's the correct answer by "writing out all of the combinations"
As usual, I really appreciate the help on these problems.
Hi jsnipes!
Unfortunately, finding this probability is actually a lot tougher that we might originally think. So what??
Okay, you were exactly right that for the first child. Sitting in front of all 4 gifts, the probability that she gets the wrong one is 3/4. Now for the second child. She is sitting in front of 3 gifts so it might seem reasonable to assume that she has a 2/3 chance of getting the wrong gift, BUT that is assuming that her correct gift is actually sitting there. Her correct gift might have been given to the first child, making the probability for the second child 3/3 of getting the wrong gift.
So let's try an easier scenario and see if we can find a pattern = 2 kids. What is the probability that at least 1 of them gets the correct gift? This is actually an OR probability = P(A or B). BUT, these are not completely independent events, because they can both get the right gift. If we simply add their probabilities, we are over-counting the times where they BOTH get the right gift.
P(A or B) = P(A) + P(B) = 1/2 + 1/2 = 1 (and that doesn't make sense!!)
Therefore, we need to subtract off the times when they are both getting the correct gift. The formula for this is the following:
P(A or B) = P(A) + P(B) - P(A & B)
= 1/2 + 1/2 - (1/2)(*1/1) = 1 - 1/2
Now let's try with 3 choices. We have to add one more piece however. If we simply subtract out the times they overlap (in pairs), we have taken too much away (we're subtracting A, B and C each twice. So it is like we need to add back in the joint probability of all 3 occurring...
P(A or B or C) = P(A) + P(B) + P(C) - P(A & B) - P(A & C) - P(B & C) + P(A & B & C)
= 1/3 + 1/3 + 1/3 - 1/3(1/2) - 1/3(1/2) - 1/3(1/2) + 1/3(1/2)(1/1)
= 1 - 3/6 + 1/6
= 1 - 1/2 + 1/6
If we do this with 4 people we get even more to add together:
P(A or B or C or D)
= P(A) + P(B) + P(C) + P(D)
- P(A & B) - P(A & C) - P(A & D)- P(B & C) - P(B & D) - P(C & D)
+ P(A & B & C) + P(A & B & D) + P(A & C & D)+ P(B & C & D)
- P(A & B & C & D)
P(A or B or C or D)
= 1/4 + 1/4 + 1/4 + 1/4
- 1/4(1/3) - 1/4(1/3) - 1/4(1/3)- 1/4(1/3) - 1/4(1/3) - 1/4(1/3)
+ 1/4(1/3)(1/2) + 1/4(1/3)(1/2) + 1/4(1/3)(1/2) + 1/4(1/3)(1/2)
- 1/4(1/3)(1/2)(1/1)
= 1 - 6/12 + 4/24 - 1/24
= 1 - 1/2 + 1/6 - 1/24
Starting to see a pattern??
It might be even easier to see if we look at the numbers 2, 6 and 24...these are actually common factorial = 2!, 3! and 4!. And now we have the formula.
If we have N grandchildren, and we want to see the probability of at least one of them getting a gift, we build the formula:
1 - 1/2! + 1/3! - 1/4! + 1/5! - ... until we get to 1/N!
For 4 grandchildren we have:
= 1 - 1/2 + 1/6 - 1/24
= (24 - 12 + 4 - 1)/24
= 15/24
= 5/8.
The correct answer is
C
Whit
