A 3-digit number is chosen at random. What is the probability that either all th

Question

A 3-digit number is chosen at random. What is the probability that either all the digits are odd or all the digits are even? Assume the first digit is NOT zero

Explanation / Answer

Think of each digit as a separate number. so we have three digits and need to know the probability of all of them being odd. BUT, since the first digit must be >0 (to make it a valid three digit number), the range for that digit is 1-9, whereas the other two digits have range 0-9 P(odd X in 1-9) = 5/9 P(odd X in 0-9) = 5/10 P(3 digit number being all odd) = (5/9)(1/2)(1/2) = 5/36 P(even X in 1-9) = 4/9 P(even X in 0-9) = 5/10 P(3 digit number being all even) = (4/9)(1/2)(1/2) = 4/36 P(3 digit number being all even or all odd) = 4/36 + 5/36 = 9/36 = 1/4 Hope that helps

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