My question refers to pp 87-88 of the Probability section in the Manhattan Guide for Word Translations (ed. 4.1).
I am having some difficulty understanding Manhattan's explanation on how to use the P(A OR B)= P(A) + P(B) - P(A AND B) formula when dealing with probability problems that feature "OR" events that CAN occur together.
The guide indicates:
If we are using the P(H1 OR H2) approach, we are looking for the number of events that will produce either {a heads on first flip with tails on second flip} OR {a tails on first flip with heads on second flip}. Therefore, we should NOT include any of the other possible events: {tails on first, tails on second}, {heads on first, heads on second}. The possible outcomes of two coin flips are: HH, HT, TH, TT; therefore, the probability of getting either H1 OR H2 is 2/4 or 1/2 (based on the HT, TH results).
Consequently, the example Manhattan cites does not seem to work here. You can't use the P(H1 OR H2) to find the probability of getting heads AT LEAST once in two fair coin flips. The phrase "at least" implies that heads, in addition to the HT or TH results, CAN actually occur on BOTH flips giving an HH result. Therefore, why would you want to subtract out the HH result (as indicated by Manhattan's (1/2) (1/2) calculation above) when it should actually be included in the probability count?
If anyone can provide some clarification on this I would really be grateful as I've been racking my brains trying to understand Manhattan's explanation.
I am having some difficulty understanding Manhattan's explanation on how to use the P(A OR B)= P(A) + P(B) - P(A AND B) formula when dealing with probability problems that feature "OR" events that CAN occur together.
The guide indicates:
But surely this doesn't make much sense.Manhattan wrote:if you simply add P(A) and P(B) in situations when A and B occur together, then you are overestimating the probability of either A or B (or both) occurring. You would be making the same error if you said that because you have a 50% chance of getting heads on one coin flip, you simply double that chance to 100% for two coin flips. But you will not always get heads at least once in two coin flips. What you have to do is subtract the chance of getting head on both flips, because the event H1 ("heads on the first flip") CAN occur together in the same scenario with event H2 ("heads on the second flip"). In other words, you CAN get two heads in a row, so you must subtract that possibility in the formula:
P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)
= 1/2 + 1/2 - (1/2) (1/2)
= 3/4
Therefore, 3/4 or 75% is the true chance of getting heads at least once in two fair coin flips.
If we are using the P(H1 OR H2) approach, we are looking for the number of events that will produce either {a heads on first flip with tails on second flip} OR {a tails on first flip with heads on second flip}. Therefore, we should NOT include any of the other possible events: {tails on first, tails on second}, {heads on first, heads on second}. The possible outcomes of two coin flips are: HH, HT, TH, TT; therefore, the probability of getting either H1 OR H2 is 2/4 or 1/2 (based on the HT, TH results).
Consequently, the example Manhattan cites does not seem to work here. You can't use the P(H1 OR H2) to find the probability of getting heads AT LEAST once in two fair coin flips. The phrase "at least" implies that heads, in addition to the HT or TH results, CAN actually occur on BOTH flips giving an HH result. Therefore, why would you want to subtract out the HH result (as indicated by Manhattan's (1/2) (1/2) calculation above) when it should actually be included in the probability count?
If anyone can provide some clarification on this I would really be grateful as I've been racking my brains trying to understand Manhattan's explanation.
, as it indicates that heads can appear in different orientations within a single event. 
