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Chapter 9: Problem 508

What is the probability of getting exactly 3 heads in 5 flips of a balanced coin?

Short Answer

Expert verified
The probability of getting exactly 3 heads in 5 flips of a balanced coin is \(0.3125\), or \(31.25\%\) using binomial probability.

Step by step solution

01

Find the number of combinations

We first need to find the number of combinations (C) of getting exactly 3 heads in 5 flips. We can use the formula for combinations: \[C(n,k) = \frac{n!}{k!(n-k)!}\] where n is the total number of flips and k is the number of heads we want to obtain. In this case, n=5 and k=3.
02

Calculate the combinations of getting exactly 3 heads

Using the formula from Step 1, we can now calculate the combinations of getting exactly 3 heads in 5 coin flips: \[C(5,3) = \frac{5!}{3!(5-3)!}\] First, calculate the factorials: 5! = 5 脳 4 脳 3 脳 2 脳 1 = 120 3! = 3 脳 2 脳 1 = 6 (5-3)! = 2! = 2 脳 1 = 2 Now plug the factorials into the formula: \[C(5,3) = \frac{120}{6 \times 2} = \frac{120}{12} = 10\] There are 10 different combinations of getting exactly 3 heads in 5 coin flips.
03

Calculate the probability of each combination

Since the coin is balanced, the probability of getting a head is 0.5, and the probability of getting a tail is also 0.5. For each of the 10 combinations from Step 2, the probability of getting exactly 3 heads and 2 tails in 5 flips is: \(P(X=3) = (0.5)^3 (0.5)^2\)
04

Calculate the probability of getting exactly 3 heads in 5 flips

To find the total probability, we need to multiply the probability of each combination by the number of combinations, which we found in Steps 2 and 3: Probability of getting exactly 3 heads in 5 flips: \(P(X=3) = 10 \times (0.5)^3 (0.5)^2\) Now, perform the calculations: \(P(X=3) = 10 \times (0.125) (0.25) = 10 \times 0.03125\) \(P(X=3) = 0.3125\) The probability of getting exactly 3 heads in 5 flips of a balanced coin is 0.3125, or 31.25%.

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Most popular questions from this chapter

Find the expected value of the random variable \(\mathrm{Y}=\mathrm{f}(\mathrm{X})\), when \(\mathrm{X}\) is a discrete random variable with probability mass function \(\mathrm{g}(\mathrm{x})\). Let \(\mathrm{f}(\mathrm{X})=\mathrm{X}^{2}+\mathrm{X}+1\) and \(\operatorname{Pr}(X=x)=g(x)=\) \(\mathrm{x}=1\) \(=\quad(1 / 3) \quad x=2\) \(=\) \(\mathrm{x}=3 .\)

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