Question

In a board game, students draw a number do not replace it, and then draw a second number. Determine the probability of each event occurring. Drawing an odd number then drawing a 6? dependent events. PLEASE HELP ME

Answer 1

The figure depicting the board game is attached below.

Answer:

Explanation:

Kindly note that selections done without replacement.

Count of numbers on the board game = 8

Count of odd numbers = (1, 9, 1) = 3

Count of digit 6 = 3

Probability = required outcome / Total possible outcomes

P(picking an odd number) = 3 / 8

Without replacement

Numbers left on board game = 8 - 1 = 7

P(picking a 6) = 3 / 7

Hence,

P(picking an odd number then, a 6) = 3/8 * 3/7 = 9 / 56