1 Expert Answer
the probability that an ace is drawn is 4/52, or 1/13 the probability that a heart is drawn is 13/52, or 1/4 don't forget that one of the aces is the ace of hearts P(an ace or heart is drawn)=4/52+13/52-1/52 or look at it this way... P=13/52+3/52=16/52 because one of the aces, the ace of hearts, is already include in the 13/52 you add the probabilities because of the word "or" the odds (in favor) of an ace or a heart is the ratio of the probability that the event will occur to the probability that the event will not occur odds=(16/52)/(36/52)=16/36=4/9, or simply look at 4/13, 13-4=9, and the ratio is 4/9
the event cannot happen in 36 ways: 13 diamonds+13 spades+13 clubs-ace of diamonds-ace of spades-ace of clubs=39-3=36, so 16(ways it can happen)/36(ways it cannot happen)=16/36=4/9
note: don't be mislead looking at 4/9, it is not probability but odds; the event can happen in 4 ways and cannot happen in 9 ways, so the event can happen in 4 ways out of a total of 4+9=13 ways
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Please let me know if my reasoning is correct: Assuming no replacement, the event space of drawing a heart first and then an ace is partitioned into the following disjoint events:
- Ace of heart, Non-heart ace (Probability = $\frac{1}{52}.\frac{3}{51}$)
- Heart but not ace, ace of heart (Probability = $\frac{12}{52}.\frac{1}{51}$)
- Heart but not ace, Non-heart ace (Probability = $\frac{12}{52}.\frac{3}{51}$)
So the probability of drawing a heart first and then an ace is the sum of the probabilities of the 3 events.
Sum of events $1, 2, 3$ is $\frac{51}{(52)(51)} = \frac{1}{52}$ So this is the probability of drawing a head first and then an ace...
Is this correct?
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