"A boy flips 4 pennies simultaneously and records the total number of heads that come up. How many outcomes are possible in the sample space of this activity?"
I initially overlooked the word "simultaneously" in this problem and chose 16, based on the Counting Principle. I had a feeling that I got the problem wrong, because it was too easy, but was not sure what else to choose. I've since learned that the correct answer is 5, and after some googling, I understand that when outcomes are indistinguishable in the sample space (because they are non-unique and part of the same event) then they are "collapsed". So, all heads (HHHH) is one outcome, all combinations of 3 heads are collapsed into one outcome (HHHT, HHTH, HTHH, THHH), all combinations of 2 heads are collapsed into one outcome (HHTT, HTHT, HTTH, THHT, THTH, TTHH), all combinations of 1 heads are collapsed, and then no heads TTTT. Once collapsed there are 5 outcomes in the sample space. OK. I get this and could answer similar problems correctly.
My question relates to how this effects calculating probability in general. Let's use a less cumbersome example of two coins tossed. If tossed sequentially there would be 4 outcomes in the sample space (HH, HT, TH, TT) and if we were to calculate the probability of getting heads exactly once it would be 2/4 or 50%. However, if the coins are tossed simultaneously, and they are otherwise indistinguishable, HT and TH collapse and the sample space would have a count of 3 (HH, HT/TH, TT). What is a good way of thinking how to calculate probability when the sample space is collapsed like this? The odds of getting heads should not change whether the coins are tossed sequentially or simultaneously.
Thanks for your help!
I initially overlooked the word "simultaneously" in this problem and chose 16, based on the Counting Principle. I had a feeling that I got the problem wrong, because it was too easy, but was not sure what else to choose. I've since learned that the correct answer is 5, and after some googling, I understand that when outcomes are indistinguishable in the sample space (because they are non-unique and part of the same event) then they are "collapsed". So, all heads (HHHH) is one outcome, all combinations of 3 heads are collapsed into one outcome (HHHT, HHTH, HTHH, THHH), all combinations of 2 heads are collapsed into one outcome (HHTT, HTHT, HTTH, THHT, THTH, TTHH), all combinations of 1 heads are collapsed, and then no heads TTTT. Once collapsed there are 5 outcomes in the sample space. OK. I get this and could answer similar problems correctly.
My question relates to how this effects calculating probability in general. Let's use a less cumbersome example of two coins tossed. If tossed sequentially there would be 4 outcomes in the sample space (HH, HT, TH, TT) and if we were to calculate the probability of getting heads exactly once it would be 2/4 or 50%. However, if the coins are tossed simultaneously, and they are otherwise indistinguishable, HT and TH collapse and the sample space would have a count of 3 (HH, HT/TH, TT). What is a good way of thinking how to calculate probability when the sample space is collapsed like this? The odds of getting heads should not change whether the coins are tossed sequentially or simultaneously.
Thanks for your help!
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