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Over the years, the percentage of candidates passing an entrance exam to a prestigious law school is \(20 \%\). At one of the testing centers, a group of 50 candidates take the exam and 20 pass. Is this odd? Answer on the basis that \(X \geq 20\) where \(X\) is the number that pass in a group of 50 when the probability of a pass is \(0.2\).

Short Answer

Expert verified
To ascertain if it's odd that 20 out of 50 candidates pass, calculate and interpret the probability \(P(X \geq 20)\). This is done effectively by using the binomial probability formula and it's complementary rule.

Step by step solution

01

Understanding The Binomial Probability Formula

The binomial probability formula is given as \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\) where \(P(X=k)\) is the probability of k successes in n trials, \(C(n, k)\) is the combinations of n items taken k at a time, \(p\) is the probability of success, \(n\) is the number of trials, and \(k\) is the number of successes. In this scenario, \(n=50\), \(p=0.2\), and the task is to find \(P(X \geq 20)\).
02

Calculate the Probability for \(X \geq 20\)

To find \(P(X \geq 20)\), one must compute the sum of the individual probabilities from \(20\) to \(50\) inclusive. This involves substituting for \(n, p\) and \(k\) from \(20\) to \(50\) in the binomial probability formula and summing up.
03

Use Complementary Rule

Notice that manually calculating each individual binomial probability for \(k\) from \(20\) to \(50\) can be very tedious. Utilize the fact that for any event \(E\), \(P(E) = 1 - P(E')\), where \(E'\) is the complement of event \(E\). This yields \(P(X \geq 20) = 1 - P(X < 20) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 19))\). The computation is much simpler here.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
To understand the binomial distribution, imagine flipping a coin. We know that getting heads or tails is a random event, but if we flip the coin multiple times, we start to see a pattern of how many times heads is likely to come up. Similarly, the binomial distribution helps us predict the probability of a certain number of successes in a fixed number of trials, like passing an exam, given that the chance of success in each trial is the same.

The formula for the binomial probability is \(P(X=k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{(n-k)})\), where:
  • \(P(X=k)\) represents the probability of having exactly \(k\) successes,
  • \(C(n, k)\) is the number of combinations of \(n\) items taken \(k\) at a time,
  • \(p\) is the probability of success on a single trial, and
  • \(n\) is the total number of trials.

Using the binomial formula, we can address questions related to patterns in random events where there are only two outcomes, such as pass or fail, on each trial.
Probability Theory
Probability theory is the branch of mathematics focused on quantifying random events. It lets us make statements about the likelihood of various outcomes, allowing for informed predictions and decisions. The core idea of probability is quite intuitive: toss a dice, and you can calculate the probability of rolling a four. In more complex situations, like predicting weather or market trends, the underlying principle remains the same, albeit with more elaborate models and computations.

In the context of the binomial distribution, probability theory provides a structured way of calculating the likelihood of a given number of successes across several attempts. It forms the backbone of analyzing scenarios with clear-cut outcomes and is critical in fields ranging from finance to pharmacology.
Combinatorics
Combinatorics is essentially the study of counting. This field of mathematics helps us figure out how many ways a given set of events can occur. For example, if you're ordering a pizza and you can choose from four toppings, combinatorics is what helps calculate how many possible topping combinations there are.

In relation to binomial probability, combinatorics is critical because it provides the formula, \(C(n, k) = \frac{n!}{k!(n-k)!}\), to determine the number of possible ways to achieve \(k\) successes in \(n\) trials, which is part of the binomial probability formula. Understanding this helps to clarify not just the total probability, but also how the different possible outcomes contribute to it.
Complementary Probability
Complementary probability refers to the probability that an event will not occur. It's a useful concept because it often simplifies calculations. Instead of adding up the probabilities of multiple events happening, we can subtract the probability of them not happening from 1.

For example, if we're trying to find the likelihood of at least one event occurring out of several possibilities, it might be easier to calculate the chance of none of them occurring, and then use complementary probability. So, to answer the question about the law school entrance exam, instead of calculating the probability of 20 or more candidates passing, which is complex, we compute the probability of fewer than 20 passing, and subtract this from 1. This application makes handling otherwise unwieldy probability questions manageable.

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Most popular questions from this chapter

Continuing with Exercise \(3.2 .8\), make a page of four overlay plots for the following 4 Poisson and binomial combinations: \(\lambda=2, p=0.02 ; \lambda=10, p=0.10\); \(\lambda=30, p=0.30 ; \lambda=50, p=0.50 .\) Use \(n=100\) in each situation. Plot the subset of the binomial range that is between \(n p \pm \sqrt{n p(1-p)} .\) For each situation, comment on the goodness of the Poisson approximation to the binomial.

Let \(X\) have an exponential distribution. (a) For \(x>0\) and \(y>0\), show that $$ P(X>x+y \mid X>x)=P(X>y) $$ Hence, the exponential distribution has the memoryless property. Recall from Exercise 3.1.9 that the discrete geometric distribution has a similar property. (b) Let \(F(x)\) be the cdf of a continuous random variable \(Y\). Assume that \(F(0)=0\) and \(00\). Suppose property (3.3.11) holds for \(Y\). Show that \(F_{Y}(y)=1-e^{-\lambda y}\) for \(y>0\). Hint: Show that \(g(y)=1-F_{Y}(y)\) satisfies the equation $$ g(y+z)=g(y) g(z) $$

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than \(0.99 .\) Find the smallest value of the mean that the distribution can take.

Let \(X_{1}, X_{2}, \ldots, X_{k-1}\) have a multinomial distribution. (a) Find the mgf of \(X_{2}, X_{3}, \ldots, X_{k-1}\). (b) What is the pmf of \(X_{2}, X_{3}, \ldots, X_{k-1} ?\) (c) Determine the conditional pmf of \(X_{1}\) given that \(X_{2}=x_{2}, \ldots, X_{k-1}=x_{k-1}\). (d) What is the conditional expectation \(E\left(X_{1} \mid x_{2}, \ldots, x_{k-1}\right) ?\)

If the random variable \(X\) has a Poisson distribution such that \(P(X=1)=\) \(P(X=2)\), find \(P(X=4)\).

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