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\(X\) is a binomial random variable with parameters \(n=5, p=0.3^{\wedge}-\). Use the table inChapter 12 "Appendix" to compute the probability indicated. a. \(P(X \leq 2)\) b. \(P(X \geq 2)\) c. \( P(2)\) d. \(P(0)\) e. \(P(5)\)

Short Answer

Expert verified
Use the binomial formula to compute individual probabilities; sum for ranges.

Step by step solution

01

Understand the Problem

The problem requires us to compute various probabilities for a binomial random variable \(X\) with parameters \(n=5\) and \(p=0.3\) using a probability distribution table from the Appendix. The probabilities to find are \(P(X \leq 2)\), \(P(X \geq 2)\), \(P(2)\), \(P(0)\), and \(P(5)\).
02

Compute P(X = 0, 1, 2, 3, 4, 5)

Using the binomial probability formula \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\), we will compute each necessary probability: - \(P(0) = \binom{5}{0} (0.3)^0 (0.7)^5\)- \(P(1) = \binom{5}{1} (0.3)^1 (0.7)^4\)- \(P(2) = \binom{5}{2} (0.3)^2 (0.7)^3\)- \(P(3) = \binom{5}{3} (0.3)^3 (0.7)^2\)- \(P(4) = \binom{5}{4} (0.3)^4 (0.7)^1\)- \(P(5) = \binom{5}{5} (0.3)^5 (0.7)^0\)Look these values up in the probability distribution table if available.
03

Compute P(X ≤ 2)

To find \(P(X \leq 2)\), sum the probabilities of occurring for \(X=0\), \(X=1\), and \(X=2\). Use the probabilities computed from the previous step:\[P(X \leq 2) = P(0) + P(1) + P(2)\].
04

Compute P(X ≥ 2)

To find \(P(X \geq 2)\), use the complementary probability principle. This is equal to 1 minus the probability of the complement (\(X < 2\)), so compute:\[P(X \geq 2) = 1 - P(X < 2) = 1 - (P(0) + P(1))\].
05

Record Direct Results for Given Values

Based on the probability computation: - \(P(2)\) is found directly using the formula for \(P(X=2)\). - \(P(0)\) and \(P(5)\) are also computed directly for the respective values.

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Key Concepts

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

Probability Distribution Table
A probability distribution table is a tool used to organize the probability of each possible value of a random variable in a given scenario. It provides a compact and easy-to-read summary of all probabilities that helps users quickly refer to or calculate probabilities. For a binomial random variable like in our exercise, the table lists probabilities for different values of "k" where "k" represents the number of successes in "n" trials with a given probability "p".

For example, in a scenario with a binomial random variable characterized by parameters, such as number of trials, "n=5", and probability of success in each trial, "p=0.3", the table will display results from the binomial probability formula. This often involves calculating values using combinations and powers, a systematic way to represent different outcomes of a binomial experiment.

Using the probability distribution table not only simplifies the computation of individual probabilities like P(2) but also entire ranges, such as P(X ≤ 2), by summing up the relevant results shown in the table.
Complementary Probability
Complementary probability is a concept that simplifies solving probability problems by using the fact that all probabilities in a system must sum up to 1. This means if you want to find the probability of an event happening, you can alternatively find the probability of it not happening and subtract from 1.

Consider our task to find P(X ≥ 2). Instead of calculating the probabilities for X = 2, 3, 4, and 5 separately, we use complementary probability. Find the probability of the event not happening, i.e., P(X < 2), which is easier to calculate directly as it consists of fewer values to sum (P(0) + P(1)). Then subtract this sum from 1: P(X ≥ 2) = 1 - P(X < 2).

This approach leverages the simplicity of subtracting probabilities and often reduces computational work, especially when dealing with large datasets or many potential outcomes.
Binomial Random Variable
A binomial random variable represents the number of successes in a sequence of independent experiments, or trials, each of which results in a success with probability "p" or a failure otherwise. The two parameters that define a binomial random variable are "n" (the number of trials) and "p" (the probability of success on an individual trial).

In our example, the variable X reflects this setup with 5 trials (n=5) and a success probability of 0.3 (p=0.3). Each trial is independent, meaning the outcome of one trial doesn't affect others.

The formula for the probability of exactly "k" successes is given by: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]This formula involves the binomial coefficient, denoted by \( \binom{n}{k} \), which calculates the number of ways to choose "k" successes in "n" trials, multiplying by the probability of exactly "k" successes \(p^k\), and "n-k" failures \( (1-p)^{n-k} \).

Understanding how to work with binomial random variables is key in fields like statistics and probability theory, where predicting the likelihood of an event is critical.

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Most popular questions from this chapter

About \(12 \%\) of all individuals write with their left hands. A class of 130 students meets in a classroom with 130 individual desks, exactly 14 of which are constructed for people who write with their left hands. Find the probability that exactly 14 of the students enrolled in the class write with their left hands.

An English-speaking tourist visits a country in which \(30 \%\) of the population speaks English. He needs to ask someone directions. a. Find the probability that the first person he encounters will be able to speak English. b. The tourist sees four local people standing at a bus stop. Find the probability that at least one of them will be able to speak English.

A fair coin is tossed repeatedly until either it lands heads or a total of five tosses have been made, whichever comes first. Let \(X\) denote the number of tosses made. a. Construct the probability distribution for \(X\). b. Compute the mean \(\mu\) of \(X\). c. Compute the standard deviation \(\sigma\) of \(X\).

The probability that an egg in a retail package is cracked or broken is \(0.025 .\) a. Find the probability that a carton of one dozen eggs contains no eggs that are either cracked or broken. b. Find the probability that a carton of one dozen eggs has (i) at least one that is either cracked or broken; (ii) at least two that are cracked or broken. c. Find the average number of cracked or broken eggs in one dozen cartons.

A roulette wheel has 38 slots. Thirty-six slots are numbered from 1 to \(36 ;\) the remaining two slots are numbered 0 and 00 . Suppose the "number" 00 is considered not to be even, but the number 0 is still even. In a \(\$ 1\) bet on even, the bettor pays \(\$ 1\) to play. If the ball lands in an even numbered slot, he receives back the dollar he bet plus an additional dollar. If the ball does not land on an even numbered slot, he loses his dollar. Let \(X\) denote the net gain to the bettor on one play of the game. a. Construct the probability distribution of \(X\). b. Compute the expected value \(E(x)\) of \(X,\) and explain why this game is not offered in a casino (where 0 is not considered even). c. Compute the standard deviation of \(X\).

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