Question

There are a total of 8 tiles in the bag containing the

following letters: G, I, R, V, I, A, I and N.
a) What is the probability of selecting an I, keeping it, then an A?
The probability of selecting an I is

b) Does the probability of selecting an A increase or decrease after the first “I” tile is pulled from the bag?
c) What is the probability of drawing all three “I” tiles from the bag, one at a time?

d) Will drawing an “A,” keeping it, then drawing an “I” be more or less likely than drawing an “I,” keeping it, then drawing another “I”?

Answer 1

a) P(IA) = 3/56; P(I) = 3/8

b) increases

c) 1/56

d) less

Explanation:

You want various probabilities related to drawing letters from a bag containing the letters G, I, R, V, I, A, I, and N.

The probability of drawing a given letter is the ratio of the number of that letter in the bag to the total number of letters in the bag.

a) P(IA) without replacement

There are 3 letters I in the bag. There is only one of each of the other 5 letters.

P(I) = 3/8

P(A after I) = 1/7

The probability of a sequence of I then A is the product of these probabilities:

P(IA) = (3/8)(1/7) = 3/56

b) Change in P(A)

Before the first I is pulled from the bag, P(A) = 1/8.

After the first I is pulled from the bag, P(A) = 1/7, an increase in the probability.

(You can see that the probability of selecting an A increases after each non-A letter is drawn. Ultimately, after the other 7 letters are drawn, P(A) = 1, as it is the only one left.)

c) P(III)

P(I) = 3/8 on the first draw

P(I) = 2/7 after the first I is drawn

P(I) = 1/6 after the second I is drawn

P(III) = (3/8)(2/7)(1/6) = 1/56

d) P(AI) vs P(II)

P(AI) = (1/8)(3/7) = 3/56

P(II) = (3/8)(2/7) = 6/56 = 3/28

Drawing A then I is less likely than drawing I then I.