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How do I describe events from the following exercises?

There are 2 ways to describe probabilities of events occurring: theoretical and experimental. Can you please tell me how to determine 1) the theoretical probability and 2) the experimental probability of the following 2 events described? (I am asking for 4 answers.)

Consider a regular shuffled deck of 52 cards (without Jokers)
Jack, Queen, and King have the value of 11, 12, and 13, respectively. Ace has the value 1.

1) Drawing 4 diamonds, one after another, if the cards are returned to the deck after each pick.
2) Drawing 3 cards with the same value one after the other without replacing them,

844 day(s) ago

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Answers (3)

matherapist 		Dr Bob,

I've noticed recently that you seem to think questions come from lazy students. I agree with you in principal, so i just delete the questions when they are suspect.

I have seen and answered a couple from you where you give the answer but can't figure how to get there. I've wondered about your reason, but assumed you were looking for a different approach to show students. Was I wrong?

Matherapist



Posted 844 dy ago

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matherapist 		1) Draw 4 diamonds in succession with replacement:
Theoretical: There are 13 diamonds, and 52 cards so the probability of drawing a diamond on the first try is 13/52=1/4. To do this 3 more times, the probability for each draw of a diamond with replacement is still 1/4. The probabilities multiply so to draw 4 diamonds in succession we have 1/4 x 1/4 x1/4 x1/4 =1/256.because each draw is independent of the previous one.

Experimentally, you would have to draw cards a huge number of times to approach this value.
If you are teaching, it can be simulated by using a coin and seeing the probability for getting 3 or 4 heads in a row, (1/8 and 1/16 respectively) or using 3 or 4 coins at once and throwing them in the air simultaneously will give similar results since each outcome is independent of what went before. or use 2 or 3 dice to simulate the theoretical distribution of 2-12 or 18. In these cases it is unlikely you will ever match the theoretical values exactly, because if you did match them, one more toss would give a deviation.

2 drawing without replacement.
The number of cards will be reduced by one on each draw.
The probability of drawing a card with a number on it is 1 for the first draw. we don't care what it is. If we specify the value then the probability is 4/52 =1/13.
To match the first card, there are 3 cards of that rank left in the 51 remaining cards so the probability of drawing a second one is 3/51=1/17.
For the 3rd draw there are 2 cards of the needed rank left and 50 cards total so the probability is 2/50=1/25.
Again each draw is independent so we multiply probabilities and get 1 x 1/17 x 1/25= 1/425
if we specify the first card's rank then we have 1/13 X1/425 = 1/5525

To do this experimentally, use a deck with 4 each of 3 ranks( A,2,3) and and the probability of drawing 3 in a row of an undetermined card is 1 x 3/11 x2/10 or 3/55.

Trying to show probabilities experimentally is a very tedious undertaking.

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Doctor Bob 		What have you tried so far to get your own answer?
I ask because I wish to help but NOT do it for you.
Dr. Bob

Posted 844 dy ago

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