Question

You are choosing two cards, replacing the first card in the deck after it has been drawn. What is the probability you choose…17.a 7 then a 3?18.two consecutive fours?19.two consecutive diamonds?

Answer 1

We will have the following:

Assuming we are talking about a standard deck of cards, then:

17. The probability of getting a 7 and then a 3 (with replacement) will be given as follows:

We know that there will be 4 cards that would be a 7 (One for each suit) and 4 cards that would be a 3 (One for ach suit); so the probability of drawing either of them in a one by one case is the same 4/52, then the probability of drawing one after the another will be:

`\begin{gathered} P=(4)/(42)\ast(4)/(52)\Rightarrow P=(1)/(169) \\ \\ \Rightarrow P\approx0.0059 \end{gathered}`

So, the probability is exactly 1/169, that is approximately 0.0059. (0.59%).

18. The probability of two consecutive fours will follow the same principle as before, and the calculations will be given by the same values. There are 4 cards that correspond to a "4" for each suit, so the probability of drawing a 4 is 4/52; and since we replace the 4 then probability of drawing a 4 again will be 4/52; thus:

The probability is exactly 1/169, that is approximately 0.0059. (0.59%).

19. The probability of drawing a diamond on the other hand is different; there are 13 cards for the diamond suit; so the probability of drawing a diamond is 13/52; and thus the probability of drawing 2 diamonds in a row is:

`\begin{gathered} P=(13)/(52)\ast(13)/(52)\Rightarrow P=(1)/(16) \\ \\ \Rightarrow P=0.0625 \end{gathered}`

So, the probability of drawing two diamonds in a row is 0.0625, (6.25%).