/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 444 A die is tossed five times. What... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 444

A die is tossed five times. What is the probability that an ace will appear: (a) at least twice; (b) at least once?

Short Answer

Expert verified
The probability of getting an ace at least twice in five tosses of a fair die is approximately 0.1962, whereas the probability of getting an ace at least once is approximately 0.5981.

Step by step solution

01

Determine the probability of getting an ace in one toss

There are 6 sides on a fair die, and 1 of them is an ace. Therefore, the probability of getting an ace in one toss is \(\frac{1}{6}\).
02

Probability of getting an ace at least twice (Scenario A)

To find the probability of getting an ace at least twice, we can use the complement rule. The complement of "getting an ace at least twice" is "getting an ace 0 or 1 times". We can find the probability of the complement and subtract it from 1 to get the probability we want. For Scenario A, the probability of getting an ace at least twice is: \(P(X \ge 2) = 1 - [P (X=0) + P (X=1)]\) Now, we'll calculate the probabilities of getting an ace 0 times and 1 time using the binomial probability formula.
03

Calculate Probabilities for Scenario A

Get ace 0 times: \(P(X = 0) = C(5, 0) * (\frac{1}{6})^0 * (\frac{5}{6})^5 = 1 * 1 * (\frac{5}{6})^5 \approx 0.4019\) Get ace 1 time: \(P(X = 1) = C(5, 1) * (\frac{1}{6})^1 * (\frac{5}{6})^4 = 5 * \frac{1}{6} * (\frac{5}{6})^4 \approx 0.4019\) Now, substituting these probabilities into the previous formula for Scenario A: \(P(X \ge 2) = 1 - [P (X=0) + P (X=1)] = 1 - [0.4019 + 0.4019] = 1 - 0.8038 \approx 0.1962\)
04

Probability of getting an ace at least once (Scenario B)

For Scenario B, the probability of getting an ace at least once is the complement of "getting an ace 0 times" which is: \(P(X \ge 1) = 1 - P (X=0)\)
05

Calculate Probability for Scenario B

Using the probability of getting an ace 0 times from Step 3: \(P(X \ge 1) = 1 - P (X=0) = 1 - 0.4019 = 0.5981\) So, the probability of getting an ace at least once is approximately 0.5981. #Conclusion# The probability of getting an ace at least twice in five tosses of a fair die is approximately 0.1962, whereas the probability of getting an ace at least once is approximately 0.5981.

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