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Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf

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Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2 Solution Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2.

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Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2 Solution Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2

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Pr(E) shows the probability of only event E happening.Pr(F) shows .pdf

  • 1. Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2 Solution Pr(E) shows the probability of only event E happening. Pr(F) shows the probability of only event F happening. Pr(EUF) shows the probability of happening of either one of them, which would be, only E happening + only F happening, - both E and F happenig at the same time(which is Pr(EnF). Reason for subtractinig Pr(EnF) is, because these are common elements, and hence are already counted twice in only E happening, and only F happening, so we had to subtract it once. Thus Pr(EUF) = Pr(E) + Pr(F) - Pr(EnF) putting values, 0.7 = 0.3 + 0.6 - Pr(EnF) which gives, Pr(EnF) = 0.2