Question

rita rolls a standard, 6 - sided die. let a be the event that the number she rolls is even. let b be the event that the number she rolls is less than 6. complete the paragraph. the conditional probability of b given a is the probability that the number rita rolls is. the conditional probability of b given a is.

Explanation:

Step1: Determine event A and B

Event A (even number): {2, 4, 6} (3 outcomes). Event B (less than 6): {1, 2, 3, 4, 5} (5 outcomes). The intersection \( A \cap B \): {2, 4} (2 outcomes, since 6 is not less than 6).

Step2: Calculate conditional probability \( P(B|A) \)

The formula for conditional probability is \( P(B|A)=\frac{P(A\cap B)}{P(A)} \). Since \( P(A\cap B)=\frac{2}{6} \) and \( P(A)=\frac{3}{6} \), then \( P(B|A)=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3} \). Also, the conditional probability of B given A is the probability that the number is less than 6, given it's even. The even numbers are 2,4,6; among these, less than 6 are 2,4 (so "less than 6, given it is even" or more precisely, the description is "less than 6" under the condition that it's even, so the first blank is "less than 6 (given that it is even)" but in context, the first part is the description: the number is less than 6 (since we are given A, so within even numbers, we check less than 6).

Answer:

The conditional probability of \( B \) given \( A \) is the probability that the number Rita rolls is \(\boldsymbol{\text{less than 6 (and even)}}\) (or more precisely, "less than 6" given it's even; the first blank's context: the number is less than 6, with the condition of being even already applied). The conditional probability of \( B \) given \( A \) is \(\boldsymbol{\frac{2}{3}}\).

(For the first blank, the intended description is that the number is less than 6 (since we are in the context of event A, so the event is "less than 6" when we know it's even). So filling the first blank with "less than 6" and the second with \(\frac{2}{3}\).)