Suppose that from a standard deck, you draw three cards without replacement. What is the expected number of spades that you will draw

Answer: The expected number of spades that you will draw is 0.751 spades

Step-by-step explanation:

The expected value can be calculated as:

∑xₙ*pₙ

Where xₙ is the n-th event, and pₙ is the probability of that event.

First, let's count the possible events and calculate the probability for each one.

x₀ = drawing 0 spades.

Out of 52 cards, we have only 13 spades, then 52 - 13 = 39 are not spades.

Then the probability of not drawing a spade in the first draw is:

p1 = 39/52

In the second draw we will have a card less than before in the deck (so we have 38 cards that are not spades, and 51 cards in total), then the probability of not drawing a spade is:

p2 = 38/51

And with the same reasoning, in the third draw the probability is:

p3 = 37/50

The joint probability for this event will be:

p₀ = p1*p2*p3 = (39/52)*(38/51)*(37/50) = 0.413

Second event:

x₁ = drawing one spade.

Let's suppose that in the first draw we get the spade, the probability will be:

p1 = 13/52

In the second draw, we get no spade, then the probability is:

p2 = 39/51

in the third draw we also get no spade, the probability is:

p3 = 38/50

And we also have the case where the spade is drawn in the second draw, and in the third draw, then we have 3 permutations, this means that the probability of drawing only one spade is:

p₁ = 3*p1*p2*p3 = 3*(13/52)(39/51)*(38/50) = 0.436

third event:

x₂ = drawing two spades:

Let's assume that in the first draw we do not get a spade, then the probabilities are:

p1 = 39/52

p2 = 13/51

p3 = 12/50

And same as before, we will have 3 permutations, because we could not draw a spade in the second draw, or in the third, then the probability for this case is:

p₂ = 3*p1*p2*p3 = 3*( 39/52)*(13/51)*(12/50) = 0.138

And the last event:

x₃ = drawing 3 spades.

The probabilities will be:

p1 = 13/52

p2 = 12/51

p3 = 11/50

And there are no permutations here, so the joint probability is:

p₃ = p1*p2*p3 = (13/52)*(12/51)*(11/50) = 0.013

Now we can calculate the expected value:

EV = 0*0.413 + 1*0.436 + 2*0.138 + 3*0.013 = 0.751

The expected number of spades that you will draw is 0.751 spades

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