If an integer is randomly selected from all positive 2-digit integers, what is the probability that the integer chosen has
(a) a 4 in the tens place?
(b) at least one 4 in the tens place or the units place?
(c) no 4 in either place?

Solution:

(a) total number of 2-digit terms =(99 - 10) + 1 = 90

n(s) = 90

number of 2 digits that have 4 in tens place = 40, 41, 42 ......49

n(a) = 10

P(A) = n(a)/n(s)

Substituting the values

P(A) = 10/90

P(A) = 1/9

Therefore, P(A) = 1/9

(b) From the previous one, we know the P(A) = 1/9

Number of 4's in units place = 14, 24, 34.....94 = 9

n(b) = 9

P(B) = n(b)/n(s)

Substituting the values

P(B) = 9/90

P(B) = 1/10

P(A and B) = 1/90

Using OR probability, we have

P(A or B) = P(A) + P(B) - P(A and B)

P(1/9 or 1/10) = 1/9 + 1/10 - 1/90

P(1/9 or 1/10) = 1/5

Therefore, P(1/9 or 1/10) = 1/5

= 1 - 1/5 = 4/5

Therefore, probability of no 4 in either place = 4/5

Therefore, the probability that the integer chosen has a 4 in the tens place is 1/9, P(1/9 or 1/10) = 1/5, and probability of no 4 in either place = 4/5.