The unbiased dice is tossed until a number greater than 4 appears. What is the probability that an even number of tosses is needed?

To find the probability that an even number of tosses is needed until a number greater than 4 appears on a fair six-sided die, we first identify the outcomes of interest. The numbers greater than 4 are 5 and 6, while the numbers 1 through 4 do not meet this criterion.

Understanding the Outcomes

When rolling a die, the probabilities are as follows:

• Probability of rolling a number greater than 4 (5 or 6): 2/6 = 1/3
• Probability of rolling a number 4 or less (1, 2, 3, or 4): 4/6 = 2/3

Calculating the Probability

We want to find the probability that the first occurrence of a number greater than 4 happens on an even toss. This can happen in the following scenarios:

• First toss: 1, 2, 3, or 4 (probability = 2/3)
• Second toss: 5 or 6 (probability = 1/3)
• First toss: 1, 2, 3, or 4 (probability = 2/3)
• Third toss: 1, 2, 3, or 4 (probability = 2/3)
• Fourth toss: 5 or 6 (probability = 1/3)

This pattern continues, and we can express the probability of needing an even number of tosses as an infinite series:

Series Representation

The probability of needing 2n tosses can be represented as:

P(even) = (2/3)^(2n-1) * (1/3)

Summing the Series

To find the total probability, we sum this series:

P(even) = (1/3) * Σ (2/3)^(2n-1) from n=1 to ∞

This is a geometric series with the first term a = (1/3) and common ratio r = (2/3)^2 = 4/9:

Using the formula for the sum of an infinite geometric series, S = a / (1 - r), we get:

S = (1/3) / (1 - 4/9) = (1/3) / (5/9) = 3/5.

Final Result

The probability that an even number of tosses is needed until a number greater than 4 appears is 3/5.