Two events are mutually exclusive if they both cannot occur together. For example, if you toss a coin , let A represent a head showing up and B represent a tail showing up. These two events are mutually exclusive. You can only have a tail or head.
To explain an independent event, pick a card from a deck of 52. The probability that it is a king is 4/52. If you put the card back and draw again, the probability is still 4/52. The second draw is independent of the first draw. If you draw another card without putting it back, its probability changes to 3/51. It becomes a dependent event.
In short, a mutually exclusive event is not an independent event.
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Yes, they are. Mutually exclusive events cannot occur together. Complementary events cannot occur together either because an event and its complement are the negative of each other.
Not necessarily. The probability of a complementary event with probability p is 1-p. Two mutually exclusive events, however, don't necessarily add up to a probability of 1. For example, the probability of drawing a King from a standard deck of cards is 1 in 13, which the complementary probability of not drawing a King is 12 in 13. The probability, however, of drawing a Heart is 1 in 4, while the probability of drawing a Club is also 1 in 4. That leaves Diamonds and Spades, which account for the remaining probability of 2 in 4.
Not quite. The listing must also be exhaustive: it must contain all possible outcomes.For the roll of a fair cubic die, consider the following:Prob(1) = 1/6Prob(2) = 1/6This is a mutually exclusive listing of the outcomes of the experiment and the corresponding probabilities of occurrence but it is not a probability distribution because it does not include all possible outcomes. As a result, the total of the listed probabilities is less than 1.
Two events are equally unlikely if the probability that they do not happen is the same for each event. And, since the probability of an event happening and not happening must add to 1, equally unlikely events are also equally likely,
If heads comes up on the first throw (p=0.5) then tails must come up on the next two throws (each has a p= 0.5 of occurring) so 0.5 x 0.5 x 0.5 = 0.125 for this event (HTT) The second event is THT and the third event TTH each has a 0.125 probability, so 3 x 0.125 = 0.375. You can also solve this problem with the binomial distribution p = 0.5 q= 0.5, p(x=1, n= 3) = 3!/(2! 1!)(0.5)^1(0.5)^2 = 0.375 If heads comes up on the first throw (p=0.5) then tails must come up on the next two throws (each has a p= 0.5 of occurring) so 0.5 x 0.5 x 0.5 = 0.125 for this event (HTT) The second event is THT and the third event TTH each has a 0.125 probability, so 3 x 0.125 = 0.375. You can also solve this problem with the binomial distribution p = 0.5 q= 0.5, p(x=1, n= 3) = 3!/(2! 1!)(0.5)^1(0.5)^2 = 0.375