Six slips of paper, with a different digit from 1 to 6 printed on each, are placed in a box. The papers are well mixed. One slip is drawn and replaced. The papers are well mixed, and a second slip is drawn. Find the probability of each of the following:

a) the first is marked with an odd number and the second with an even number (is it 1/6 and 3/6?)
b)the first is marked with a four and the second with a number less than four(is it (1/6)x(1/2)?
c)both slips are marked with an even number
Thanks in advance

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

a) 3/6 * 3/6 = ?

b) 1/6 * 1/2 = ?

c) 3/6 * 3/6 = ?

1/2

To find the probability for each scenario, we need to consider the total number of possible outcomes and the number of favorable outcomes for each case.

a) The probability of drawing an odd number on the first slip is 3/6 since there are three odd numbers (1, 3, and 5) out of the total of six slips. After replacing the slip, the probability of drawing an even number on the second slip is also 3/6 since there are three even numbers (2, 4, and 6) remaining in the box. To find the probability of both events occurring, we multiply the probabilities: (3/6) * (3/6) = 9/36 = 1/4.

b) The probability of drawing number four on the first slip is 1/6 since there is only one slip with the number four out of the total of six slips. After replacing the slip, the probability of drawing a number less than four on the second slip is 3/6 since there are three slips (1, 2, and 3) remaining in the box that satisfy this condition. To find the probability of both events occurring, we multiply the probabilities: (1/6) * (3/6) = 3/36 = 1/12.

c) The probability of drawing an even number on the first slip is 3/6 since there are three even numbers (2, 4, and 6) out of the total of six slips. After replacing the slip, the probability of drawing an even number on the second slip is also 3/6 since there are still three even numbers (2, 4, and 6) remaining in the box. To find the probability of both events occurring, we multiply the probabilities: (3/6) * (3/6) = 9/36 = 1/4.

So the probabilities are:
a) 1/4
b) 1/12
c) 1/4