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Chapter 4: Problem 130

How many times should a coin be tossed to obtain a probability equal to or greater than .9 of observing at least one head?

Short Answer

Expert verified
Answer: 4 coin tosses

Step by step solution

01

Define the probability of A

We want to find the probability of observing at least one head. This event can be represented as A. The probability of observing at least one head can be denoted as P(A).
02

Define the probability of A'

The complement of event A is not observing any heads (i.e., all tails). This event can be represented as A'. To find the probability of A', we need to know the probability of getting tails in one coin toss, which is 0.5. Since the coin tosses are independent, the probability of not observing any heads in n coin tosses can be computed as P(A') = (0.5)^n.
03

Use the complement probability formula

According to the complement probability formula, P(A) = 1 - P(A'). In this problem, we are given that P(A) 鈮� 0.9. Thus, we can set up the inequality: 1 - P(A') 鈮� 0.9.
04

Solve the inequality

First, we can rearrange the inequality to find the probability of A': P(A') 鈮� 1 - 0.9 P(A') 鈮� 0.1 Now, we substitute P(A') with the formula we found in step 2: (0.5)^n 鈮� 0.1 We can solve this inequality for n using logarithms: n * log(0.5) 鈮� log(0.1) n 鈮� log(0.1) / log(0.5) Calculating the values: n 鈮� -1 / -0.30103 n 鈮� 3.32
05

Find the smallest number of coin tosses

Since the number of coin tosses must be a whole number, we can round up the result from step 4: n 鈮� 4 So, the coin should be tossed at least 4 times to obtain a probability equal to or greater than 0.9 of observing at least one head.

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Player \(A\) has entered a golf tournament but it is not certain whether player \(B\) will enter. Player \(A\) has probability \(1 / 6\) of winning the tournament if player \(B\) enters and probability \(3 / 4\) of winning if player \(B\) does not enter the tournament. If the probability that player \(B\) enters is \(1 / 3,\) find the probability that player \(A\) wins the tournament.

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