A six-sided die (whose faces are numbered 1 through 6, as usual) is known to be counterfeit: The probability of rolling any even number is twice the probability of rolling any odd number. What is the probability that if this die is thrown twice, the first roll will be a 5 and the second roll will be a 6?

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∻Solution by Steps∻ ‖ step 1 ⋮ Determine the probabilities for rolling an odd number and an even number. ‖ ‖ step 2 ⋮ Let the probability of rolling an odd number be $ p $. Then the probability of rolling an even number is $ 2p $ since it is twice as likely. ‖ ‖ step 3 ⋮ There are 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6) on the die. The total probability must sum to 1, so $ 3p + 3(2p) = 1 $. ‖ ‖ step 4 ⋮ Solve for $ p $: $ 3p + 6p = 1 $ gives $ 9p = 1 $, so $ p = \frac{1}{9} $. ‖ ‖ step 5 ⋮ The probability of rolling a 5 (an odd number) is $ \frac{1}{9} $, and the probability of rolling a 6 (an even number) is $ 2p = \frac{2}{9} $. ‖ ‖ step 6 ⋮ The probability of rolling a 5 and then a 6 is the product of their individual probabilities: $ \frac{1}{9} \times \frac{2}{9} $. ‖ ‖ step 7 ⋮ Calculate the product: $ \frac{1}{9} \times \frac{2}{9} = \frac{2}{81} $. ‖ ∻Answer∻ ⚹ $ \frac{2}{81} $ ⚹ ∻Key Concept∻ ⚹Independent Events Probability⚹ ∻Explanation∻ ⚹The probability of two independent events occurring in sequence is the product of their individual probabilities.⚹◊What is the probability of rolling an odd number on this counterfeit die?⍭ Generate me a similar question◊

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