Asked by Kennedy
two cards are dealt in succession from a standard deck of cards. find the probability that the first card is a diamond or a king.
probability that the second card is red given that the first card was a heart.
if two dice are rolled, what is the probability of rolling at least one three?
i have 2c1(5/6)^1(1/6)^1+2c2(5/6)^1(1/6)^2 but what do i do after this. can someone please explain in regular terms, NOT math terms! thanks :)
f(x)=x+5, g(x)=x^2-2x+1
f(2)-g(2)..how do i do this????
probability that the second card is red given that the first card was a heart.
if two dice are rolled, what is the probability of rolling at least one three?
i have 2c1(5/6)^1(1/6)^1+2c2(5/6)^1(1/6)^2 but what do i do after this. can someone please explain in regular terms, NOT math terms! thanks :)
f(x)=x+5, g(x)=x^2-2x+1
f(2)-g(2)..how do i do this????
Answers
Answered by
PsyDAG
The first card has a 1/4 chance of being a diamond. Since the phrase is "or," it now excludes all diamonds, including the king of diamonds. So now the chances of getting a king are 3 kings out of the remaining 39 cards. In an "either-or" situation you add the individual probabilities of the specific events.
There is a 1/4 chance of getting a heart on the first card. Without replacement, the chances of getting a red card are 25 out of the remaining 51 cards. Since you want to know the probability of "both/all" events occurring, you multiply the individual probabilities.
"At least one three" means one three or two threes. Since each die has 6 sides, the probability = 1/6 + 1/36. ("either-or" again)
I hope this helps. Thanks for asking.
There is a 1/4 chance of getting a heart on the first card. Without replacement, the chances of getting a red card are 25 out of the remaining 51 cards. Since you want to know the probability of "both/all" events occurring, you multiply the individual probabilities.
"At least one three" means one three or two threes. Since each die has 6 sides, the probability = 1/6 + 1/36. ("either-or" again)
I hope this helps. Thanks for asking.
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