On one side of a coin there is the number 0 and on the other side the number 1. What is the probability that the sum of three coin tosses will be 2?
Should we have 8 outcomes here or 4 since the sum can only be 0, 1, 2 or 3?
Thank you ...
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Total Outcomes in Probability
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Brian@VeritasPrep
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Hey Chris,
Great question, and I think it hinges on how you interpret the term "outcomes". You're right that there are only 4 potential sums, but I'd look at "outcomes" as to define the number of sequences. The reason for that is that each sequence has the same probability, so that allows you to calculate. Since there are only 8 potential sequences we can even just list them out:
0, 0, 0 ---> sum to 0
0, 0, 1 ---> sum to 1
0, 1, 0 ---> sum to 1
1, 0, 0 ---> sum to 1
0, 1, 1 ---> sum to 2
1, 0, 1 ---> sum to 2
1, 1, 0 ---> sum to 2
1, 1, 1 ---> sum to 3
So there are 8 sequences, but only 4 sums. I'd look at "outcomes" as meaning the entire sequence and not just the sum result, because then you can calculate mathematically:
8 total sequences is your denominator
Then, calculate how many ways to get 2 as your numerator, either listing out the sequences that work, or using permutations:
How many ways to arrange two 1s and a 0?
N = 3; one repeat:
3!/2! = 3
So there are 3 sequences that add to 2 out of 8 sequences total for a probability of 3/8.
Great question, and I think it hinges on how you interpret the term "outcomes". You're right that there are only 4 potential sums, but I'd look at "outcomes" as to define the number of sequences. The reason for that is that each sequence has the same probability, so that allows you to calculate. Since there are only 8 potential sequences we can even just list them out:
0, 0, 0 ---> sum to 0
0, 0, 1 ---> sum to 1
0, 1, 0 ---> sum to 1
1, 0, 0 ---> sum to 1
0, 1, 1 ---> sum to 2
1, 0, 1 ---> sum to 2
1, 1, 0 ---> sum to 2
1, 1, 1 ---> sum to 3
So there are 8 sequences, but only 4 sums. I'd look at "outcomes" as meaning the entire sequence and not just the sum result, because then you can calculate mathematically:
8 total sequences is your denominator
Then, calculate how many ways to get 2 as your numerator, either listing out the sequences that work, or using permutations:
How many ways to arrange two 1s and a 0?
N = 3; one repeat:
3!/2! = 3
So there are 3 sequences that add to 2 out of 8 sequences total for a probability of 3/8.
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Ian Stewart
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When you calculate the 'total number of outcomes' to find the denominator of a probability, each of your outcomes must be equally likely. So in your question above, there are 8 possible outcomes in total, and not 4, since there are 8 different sequences we can have if we flip this coin three times:chrisjim5 wrote:On one side of a coin there is the number 0 and on the other side the number 1. What is the probability that the sum of three coin tosses will be 2?
Should we have 8 outcomes here or 4 since the sum can only be 0, 1, 2 or 3?
Thank you ...
000
001
010
100
110
101
011
111
Three of these give us the correct sum, so 3/8 is the answer.
It might be easier to see why this distinction is important with a different real world example. If you are asked 'what is the probability of winning the lottery', using whatever real world lottery you're familiar with, we would not plug the value 2 into our denominator because there are two outcomes (you win or you lose); those two outcomes are not equally likely (unless it's a very easy lottery to win!). Instead, you'd need to work out how many different lottery tickets are possible in total - that would be your denominator - and how many winning lottery tickets are possible in total - that would be your numerator.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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