Question :

Without looking , you take a card out of a jar and replace it before selecting again. Find the probability of each each event . Then find the probability assuming the card is not replaced after each selection.

The #'s are 5,5,6,8,7

Q #1 :

P(5,6)

[IN A PACK OF 52 PLAYING CARDS

The probability of selecting a particular card (rank and suit) is 1/52.

The probability of selecting a particular rank or number of any suit is 4/52=1/13.

The probability of selecting a court card of any suit (J, Q, K) is 12/52=3/13.

The probability of selecting a 5 then a 6 (with replacement) is (1/13)(1/13)=1/169.]

IN A JAR CONTAINING ONLY 5 CARDS
If the jar contains only 2 5s, 6, 8, 7, the chances of selecting a 5 is ⅖. The chances of then selecting the 6 is ⅕. So the combined probability is ⅖×⅕=2/25, 8% or 0.08.

 5 5 6 7 8 5 5,5 5,5 5,6 5,7 5,8 5 5,5 5,5 5,6 5,7 5,8 6 6,5 6,5 6,6 6,7 6,8 7 7,5 7,5 7,6 7,7 7,8 8 8,5 8,5 8,6 8,7 8,8

The table shows all outcomes. The numbers in the first column represent the first card drawn and the numbers in the first row represent the second card drawn. There are 25 possibilities, shown in the cells of the table. Count the number of cells containing 5,6 and you get 2 out of 25, which is the probability calculated earlier.

However, if the order is unimportant then we also need to count cells containing 6,5 as well as 5,6. This time there are 4 out of 25, giving us the probability of 4/25, 16% or 0-.16. It all depends on what you mean by P(5,6).

by Top Rated User (982k points)