Probability Question?

Question

If a probability question states that 4 cards were chosen at random from a deck, what is the chance that not all the cards are diamonds?
So when it says not all the cards are diamonds, does it mean that it can only either be 1 diamond and 3 non-diamond or 2 diamond and 2 non-diamond or 3 diamond and 1 non-diamond.. well thats what i think but my teacher insists that 4 non-diamonds is also an option.. so who's correct?

Answer 1

4 non-diamond is also an answer.

When dealing with "not" statements in probablility, P(E)=1-P(not E) (or the probablility of an event occuring is equal to 1 minus the probablity that the event does not occur).

With something like this question, either a hand falls in the question or doesn't. Obviously 4 non-diamonds are not 4 diamonds, so that means 4 non-diamonds falls in the case of not E.

Answer 2

In logic, "not all the cards are diamonds" is exactly equivalent to "one or more cards is/are non-diamonds" -- which (as your teacher says, includes the case where NON are diamonds.

There are 13 diamonsds in a deck, so the chances that the first card you draw is a diamond is 13/52. The chances that the second card is one of the remaining diamonds is 12/51. The chances that the third is a diamond as well is 11/50, and for the fourth being a diamond as well is 10/49. The chance of all four being diamonds is then 13/52 x 12/51 x 11/50 x 10/49 which is 17150 / 6497400. This means that the chances that you DON'T draw four diamonds is (6497400 - 17150) over 6497400. That is 6480250 / 6497400 that's a shade over 99.736 % chance. Getting back to the original wording, the chance that not all the cards are diamonds is 99.736 %.

Answer 3

There is a 1 in 4 chance that a single card is a diamond (if we are using a 52 card deck w/o jokers) so there is a (1/4)^4 = 1/256 chance that all the cards are diamonds, thus a 255/256 that at least one card is NOT a diamond.

However, the question may mean what is the chance that no diamonds appear? there is a 3/4 chance that a single card is not a diamond, thus a (3/4)^4 = 81/256 chance that none of the cards are diamond


...oops! I forgot about the chance of a diamond being less than 1/4 if a diamond was already picked. Semi-mad Scientist's answer looks right though.

Answer 4

I suspect my argument is not going to be well received. I say the probability is 50% Let a ∈ℝ, Let b ∈ℝ and randomly select the values for a and b. As already noted, for a ≤ 0, P( a < b²) = 1, this is trivial. Only slightly less trivial is the idea that P(a < 0 ) = 1/2 and thus P( a < b² | a ≤ 0) = 1 and P( a < b² ) ≥ 1/2 Now consider what happens when a > 0 For a > 0, while it is easy to show there is a non zero probability for a finite b, the limit, the probability is zero. a < b² is equivalent to saying 0 < a < b², remember we are only looking at a > 0. If this a finite interval on an infinite line. The probability that a is an element of this interval is zero. P( a < b² | a > 0) = 0 As such we have a total probability P( a < b² ) = P( a < b² | a ≤ 0) * P(a ≤ 0) + P( a < b² | a > 0) * P(a > 0) = 1 * 1/2 + 0 * 1/2 = 1/2 Remember, this is because of the infinite sets. No matter what type of interval you draw on paper or on a computer you will find a finite probability that appears to approach 1. But this is due to the finite random number generators on the computer and if we had this question asked with finite values there would be a a solution greater than 50%. I don't mean to be condescending, but please explain why using the Gaussian to approximate a uniform distribution is a good idea? Aren't infinite numbers fun. Cantor when mad working with them! :)

Answer 5

Its probably 1 diamond or 2 or 3 or none. The only option you CAN'T have is ALL diamonds.

Start at that point. Consider the probability that all four are diamonds and subtract that from 1.

P(no diamonds) = 1 - P(4 diamonds)

Use the combinations formulas:

P = 1 - [13choose4/52choose4]

Answer 6

The teacher is actually correct on this one. The use of "not" is logical in nature, so "not all diamonds" is the opposite of "all diamonds." Any configuration where one or more cards is a non-diamond will work.

So it could be
3 diamonds and 1 non-diamond,
2 diamonds and 2 non-diamonds,
1 diamond and 3 non-diamonds, or
4 non-diamonds.

You can also think of it like this. When we indicate a probability, we would indicate it in this form--P(D) where I am using "D" to represent "all cards are diamonds." We generally want the statement symbolized inside the parentheses to be positive--so that it doesn't have the term "not" in it. So to ask about the probability of "not all diamonds" is simply ~P(D).

So calculate the probability that all four are diamonds, then subtract from 1.

Answer 7

It is poorly worded. I would interpret "not all the cards are diamonds" means that there will be at least one non-diamond card in the 4. Like 4 of a kind (diamonds) in Poker is very rare.

So what is the chance that you will draw 4 successive cards and not get all diamonds? Pretty high. To get all diamonds the probability would be LESS than 0.25^4 because there are fewer diamonds with each draw. So to NOT get all diamonds, the probability would be MORE than 1 minus (0.25^4).

Answer 8

I think that the reason you misinterpreted the book's language is that in everyday conversation, we often say "not all" to mean "some, but not all." For example, if I say, "Not all of my cars are convertibles," you might reasonably assume that I own at least one convertible and one non-convertible. However, a logician (or a writer of math books) would not make that assumption. In fact, he/she would probably think that maybe I have no cars at all!

For the problem you wrote about:
If there are 4 cards, and we are describing them using everyday language, we might say:
1. "None are diamonds." (meaning there are 0 diamonds)
2. "Some are diamonds." (meaning there are 1, 2, or 3 diamonds)
3. "All are diamonds." (meaning there are 4 diamonds)

And we might feel that the second statement ("Some are diamonds.") is the same as "Not all are diamonds." In other words, we interpret "some" to mean "some, but not all." And we also interpret "not all" to mean "some, but not all."

But a logic (or math) professor would interpret the words differently. To a logician, "some" means "more than zero" (including all). And "not all" means "anything less than all" (including zero).

The quantitative words used in logic are: all, not all, some, and none.
Anything less than "all" is "not all" (which includes the possibility of "none").
Anything more than "none" is "some" (which includes the possibility of "all").

I hope that helps you to see where your teacher and the book's author are coming from ... and also helps you to interpret future word problems.

Answer 9

The only non-option is 4 diamonds.

Answer 10

you're teacher is right when she said that 4 non-diamonds is also an option. it's one of the differences of everyday English language and "logic problem" language.

to solve the problem more efficiently, just subtract the proby that you will draw 4 diamonds ===> P(A)

P(A)=(1/4)*(12/51)*(11/50)*(10/49)

1-P(A) will be the answer to the problem.