Probability and Stats

Question

1.     With replacement and a normal deck, no jokers, show how to calculate the probability of the following: 

a.     drawing a king and an ace on two draws

b.     drawing a 2 or a 3 on either of two draws

c.     drawing four aces in a row 

Answer 1

Hi Ryan,

Here is how you work through problems like these:

1.     With replacement and a normal deck, no jokers, show how to calculate the probability of the following: 

a.     drawing a king and an ace on two draws

This one is open to some interpretation. It doesn't specify if you a drawing a King and then and Ace, or if the order in which they are drawn doesn't matter. You will get a different answer depending on the interpretation.

If we're trying to find the probability of a King and then and Ace, we have:

P(K₁ and A₂ ) = (4/52)(4/52) = 0.0059

If the order doesn't matter, we need to multiply that answer by two, because P(K and A) = P(K₁ and A₂) or P(A₁ and K₂), and remember that in probability, "or" means "add" (and sometimes subtract, but not here!)

So we have:

P(K and A) = P(K₁ and A₂) or P(A₁ and K₂)

P(K and A) = (4/52)(4/52) + (4/52)(4/52) = 0.0118.

b.     drawing a 2 or a 3 on either of two draws

This one is even harder to interpret what is meant. I'm not sure if you are required to get both a 2 and a 3, but simply that the order doesn't matter, or if the Event is satisfied if you get either a 2 or a 3 on either one of the draws [so, for example (3 of spades, 5 of hearts) would be an outcome in the Event.] Ask your teacher about this one if you can.

c.     drawing four aces in a row 

The answer is (4/52)(4/52)(4/52)(4/52) = (4/52)⁴ = 0.000035

Note that, in part c we don't need to concern ourselves with questions about "order," since there is only one way to pick four Aces given four draws, whereas there are two ways to get a King and an Ace given two draws if we don't care about the order in which they are chosen.

I hope that all made sense!