Consider taking balls from a box without replacement.

Question

The probability of getting a red ball in the first draw is 5/8; the probability of not getting a red ball at the first draw but a red ball at the second draw is 2/3. What’s the probability of getting a red in either of the first two draws?

Answer is 7/8

Answer 1

Let P(x1,x2,...,xn) denote the probability of events x1,x2,..., and xn occurring on draws, 1,2,..., and n

-If the events are independent (meaning the individual event of drawing a ball from the box does not alter the following event), then P (A and B) = P(A)*P(B) and P(A or B) = P(A) + P(B).

-If the events are dependent (the individual event of drawing a ball from the box does alters the following event), then P(A and B) = P(A) + P(B) - P(A|B) because the individual elements for each event are inclusive (the same red balls for event A could be drawn for event B).

Similarly, P(A or B) = P(A) + P(B) - P(A and B). Note that P(A and B) is subtracted because the elements within the events are overlapped (i.e. the number of elements that could possibly be inclusive for both events). When finding a probability, only one element will be chosen.

Based on the question, the events are dependent because the ball that are drawn are not replaced to create a random event for the following draw.

The statement "What is the probability of attaining a red ball on either of the first two draws" is equivalent to saying "What is the probability of attaining a red ball on the first draw or the second draw).

Therefore, what must be solved is P(red or red). Since P(red) = 5/8, it follows that

P(red or red) = (5/8) + (4/7) - (5/8)*(4/7) = (40/56) + (28/56) -(20/56) = (48/56) or (7/8)