Statistics Question: Probability of Diamonds in a Deck

[Jamin2112]

Homework Statement

Three cards are dealt from a well-shuffled deck.

(a) Find the chance that all of the cards are diamonds.
(b) Find the chance that none of the cards are diamonds.
(b) Find the chance that the cards are not all diamonds.

Homework Equations

Not sure ...

The Attempt at a Solution

There are 52 cards in a deck. The chance of one being a diamond is 13/52. If you know that one card you were dealt is a diamond, then the chance of another being a diamond is 12/51, since there is 1 less card and 1 less diamond. And if you that two cards are diamonds, then the chance of a third being a diamond is 11/50. The chance of 3 cards being diamonds is therefore (13/52)*(12/51)*(11/50). Part (b) can be solved similarly. This time the chance is 39/52, then 38/51, then 37/50, and the chance of the three happening is (39/52)*(38/51)*(37/51). Part (c) is simply the opposite of part (a).

Am I right?

 

[QuarkCharmer]

It sounds like you have this all worked out to me. You have the probabilities for each stage correct, and multiplying the three together should lower the probability to show the probability of the two states, and then three. It looks good to me.

I came up with 11/850 for part(a) using that method.

I'm not sure that I entirely understand part(c).
"Find the chance that the cards are not all diamonds."

Sounds like this one is a little different. If the first card is a diamond, so it the second, but the third is not, it would still satisfy this condition right? I think this one is asking "What are the chances that none of the cards will be a diamond, and all of the cards will not be diamonds"

I don't think it's the opposite of part(a) as you mentioned.

 

[Jamin2112]

I'm just confused because this chapter was supposed to be about adding things that are mutually exclusive.

 

[QuarkCharmer]

I don't know what the chapter was supposed to be about, but I don't really see any other solution.

Not all the cards are diamonds...

So let's think of this in three cases.
In case 1, let's say the card cannot be a diamond
(3/4) right?

Let's also assume that because cases 1 and 2 will fulfill the "not all be diamonds" requirement, case three can be anything.
(1/1)

so what would case 2 have to be?
(3/4)(X/X)(1/1)

 

[DeeNos]

Your approach to solving parts (a) and (b) is correct. The probability of all three cards being diamonds is (13/52)*(12/51)*(11/50) = 0.002641 or approximately 0.26%. The probability of none of the cards being diamonds is (39/52)*(38/51)*(37/50) = 0.427255 or approximately 42.73%.

For part (c), the probability of the cards not all being diamonds is the complement of the probability of all three cards being diamonds. So, it would be 1 - 0.002641 = 0.997359 or approximately 99.74%. This means that there is a very high chance that at least one of the cards will not be a diamond.

In terms of equations, you can use the formula for combinations (nCr) to calculate the probability of getting a specific combination of cards. In this case, we have 13 diamonds and we want to choose 3 of them. So, the probability of getting all diamonds would be (13C3)/(52C3) = 0.002641 and the probability of getting no diamonds would be (39C3)/(52C3) = 0.427255.

Overall, your solution is correct and you have a good understanding of how to calculate probabilities in this scenario. Keep up the good work!