Question

Using both the binomial distribution and the normal approximation. Consider a biased coin with probability \(1 / 3\) of heads and \(2 / 3\) of tails and suppose it is tossed 450 times. (a) Find the probability of getting exactly 320 tails. (b) Find the probability of getting between 300 and 320 tails.

Short answer

For part (a), the probability of getting exactly 320 tails is approximated using normal distribution as \(P(1.95 < Z < 2.05)\). For part (b), the probability of getting between 300 and 320 tails is approximated as \(P(-0.05 < Z < 2.05)\).

Step-by-step solution

Define the Problem

Identify the random variable and its distribution. Here, let \(X\) be the number of tails in 450 tosses of a biased coin where each toss results in a tail with probability \(\frac{2}{3}\). \(X\) follows a binomial distribution: \[X \sim B(n=450, p=\frac{2}{3})\]

Calculate Mean and Standard Deviation

For a binomial distribution \(X \sim B(n, p)\), the mean \(\mu\) and standard deviation \(\sigma\) are given by: \[\mu = np \] \[\sigma = \sqrt{np(1-p)}\] Substitute \(n = 450\) and \(p = \frac{2}{3}\): \[\mu = 450 \frac{2}{3} = 300\] \[\sigma = \sqrt{450 \frac{2}{3} \frac{1}{3}} = \sqrt{100} = 10\]

Probability Using Binomial Distribution (a)

To find the probability of getting exactly 320 tails using the binomial distribution: \[P(X = 320) = \binom{450}{320} \left(\frac{2}{3}\right)^{320} \left(\frac{1}{3}\right)^{130}\] Calculate using a binomial probability calculator or software.

Normal Approximation (a)

When \(n\) is large, the binomial distribution \(X \sim B(n, p)\) can be approximated by a normal distribution \(Y \sim N(np, \sqrt{np(1-p)})\). Here, \(Y\) follows: \[Y \sim N(300, 10)\] To find \(P(X = 320)\), we use the continuity correction: \[P(319.5 < Y < 320.5)\] Standardize the variable: \[z_1 = \frac{319.5 - 300}{10} = 1.95\] \[z_2 = \frac{320.5 - 300}{10} = 2.05\] Find \(P(1.95 < Z < 2.05)\) using the standard normal distribution.

Probability Using Binomial Distribution (b)

For the range \(300 \leq X \leq 320\): \[P(300 \leq X \leq 320) = \sum_{x=300}^{320} \binom{450}{x} \left(\frac{2}{3}\right)^{x} \left(\frac{1}{3}\right)^{450 - x}\] Calculate using a binomial probability calculator or software.

Normal Approximation (b)

For the range \(300 \leq X \leq 320\), use normal approximation with continuity correction: \[P(299.5 < Y < 320.5)\] Standardize the variable: \[z_1 = \frac{299.5 - 300}{10} = -0.05\] \[z_2 = \frac{320.5 - 300}{10} = 2.05\] Find \(P(-0.05 < Z < 2.05)\) using the standard normal distribution.

Key concepts

biased coin probabilities

When dealing with a biased coin, the chances of landing heads or tails are not equal. In our exercise, the biased coin has a probability of landing heads at \( \frac{1}{3} \) and tails at \( \frac{2}{3} \). Understanding these probabilities is essential. Here's why: Each toss of the coin is independent of the others. So, the probability of an outcome doesn't change as we keep tossing the coin. In practical terms:

• If you toss the coin once, the chance of tails is \( \frac{2}{3} \).
• If you toss it 450 times, the law of large numbers tells us the results will average close to these probabilities. That is, we'd expect about \( 450 \times \frac{2}{3} = 300 \) tails in 450 tosses.

This calculation of tails (300) is crucial for further calculations, like finding the exact or range probabilities for different scenarios using binomial distribution.

normal approximation

The binomial distribution can become tricky to handle for large numbers. Here, our coin is tossed 450 times. To simplify, we use the normal approximation. This method turns our problem into one involving the normal distribution, making calculations straightforward.
For our binomial distribution with \( n = 450 \) and \( p = \frac{2}{3} \), we have calculated:

• Mean \( \mu = np = 300 \)
• Standard Deviation \( \sigma = \sqrt{np(1-p)} = 10 \)

Using these, we approximate our binomial distribution by a normal distribution \( N(300, 10) \). The normal approximation becomes more accurate with large sample sizes, making it useful for our task. We further use continuity correction to transition from discrete (binomial) to continuous (normal).

binomial distribution calculations

For our first task, finding exactly 320 tails, we use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Plugging our values in:
\[ P(X = 320) = \binom{450}{320} \left( \frac{2}{3} \right)^{320} \left( \frac{1}{3} \right)^{130} \] This equation can be solved using a binomial probability calculator or software, due to its complexity.
For the range probability of 300 to 320 tails: \[ P(300 \leq X \leq 320) = \sum_{x=300}^{320} \binom{450}{x} \left( \frac{2}{3} \right)^{x} \left( \frac{1}{3} \right)^{450-x} \] This also typically requires computational assistance. These calculations rely heavily on understanding and applying the binomial theorem.

standard normal distribution

To employ the normal approximation, we convert our original problem to the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This involves 'standardizing' our variable:

For finding exactly 320 tails using normal approximation:

• Apply continuity correction:
  \[ P(319.5 < Y < 320.5) \]
• Standardize:
  \[ z_1 = \frac{319.5 - 300}{10} = 1.95 \]
  \[ z_2 = \frac{320.5 - 300}{10} = 2.05 \]

Use the standard normal distribution to find \[ P(1.95 < Z < 2.05) \]
For the range 300 to 320 tails:

• Apply continuity correction:
  \[ P(299.5 < Y < 320.5) \]
• Standardize:
  \[ z_1 = \frac{299.5 - 300}{10} = -0.05 \]
  \[ z_2 = \frac{320.5 - 300}{10} = 2.05 \]

Use the standard normal distribution to find \[ P(-0.05 < Z < 2.05) \]. Calculations with the standard normal distribution offer a simpler way to find the required probabilities.