Skip to main content
Mathematics

Return to Question

added 59 characters in body
Source Link
user822241
user822241

$1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

  1. All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.

  1. Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.

  2. Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.

  3. There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).

  4. All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.

Are my solutions correct? If not, where did I go wrong?

$1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

  1. All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.

  1. Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.

  2. Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.

  3. There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).

  4. All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.

$1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

  1. All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.

  1. Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.

  2. Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.

  3. There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).

  4. All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.

Are my solutions correct? If not, where did I go wrong?

Source Link
user822241
user822241

Computing card probabilities

$1$ to $13$ and whose colors are red, green, blue, or yellow. You draw the topmost three cards. What's the probability that:

  1. All three cards are green or blue. I think the answer is ${26\choose 3}/{52\choose 3}$ since there are $26$ green or blue cards, and we pick three.

  1. Exactly two of the cards are colored blue. I think the answer is $({13\choose 2}\cdot 39)/{52\choose 3}$ since we pick the blue cards in ${13\choose 2}$ ways and pick one of the last $39$ (non-blue) cards in $39$ ways.

  2. Exactly one card has the number $13$ written on it. I think the answer is ${4\choose 1}{48\choose 2}/{52\choose 3}$ since you pick the $13$ in four ways and pick the other two cards in ${48\choose 2}$ ways.

  3. There is at least one card with a $12$ written on it, but no cards with a $13$ written on it. I think the answer is $4\cdot {47\choose 2}/{52\choose 3}$ since we can place a $12$ in four ways and pick the last $2$ cards in ${47\choose 2}$ ways (remove the one ace and four $13$'s).

  4. All of the cards are different colors. I think the answer is ${4\choose 3}13^{3}/{52\choose 3}$ since you pick the three colors in ${4\choose 3}$ ways and for each color you can permute the value in $13$ ways.