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elena-14-01-66 [18.8K]
2 years ago
13

In a board game, students draw a number do not replace it, and then draw a second number. Determine the probability of each even

t occurring. Drawing an odd number then drawing a 6? dependent events. PLEASE HELP ME
Mathematics
1 answer:
murzikaleks [220]2 years ago
7 0

The figure depicting the board game is attached below.

Answer:

Step-by-step 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

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Explanation :

Step-by-step explanation:

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<span>the complete question in the attached figure
</span>
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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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