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juan rolls a fair number cube twice what is the probability that he will roll an odd number on the first roll and an even number in second

Sagot :

semsee45

Answer:

1/4

Step-by-step explanation:

[tex]\mathsf{Probability \ of \ an \ event \ occurring = \dfrac{Number \ of \ ways \ it \ can \ occur}{Total \ number \ of \ outcomes}}[/tex]

Assuming the number cube has faces number 1 through 6, the probability of rolling any of the number is 1/6.

Therefore,

P(odd number) = P(1) or P(3) or P(5)

                         = 1/6 + 1/6 + 1/6

                         = 3/6

                         = 1/2

Similarly,

P(even number) = P(2) or P(4) or P(6)

                           = 1/6 + 1/6 + 1/6

                           = 3/6

                           = 1/2

⇒ P(odd) AND P(even) = 1/2 × 1/2 = 1/4

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