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Chapter 4: Problem 24

Eight prizes are distributed by a lottery. The first participant takes 5 tickets from the box containing 50 tickets. What is probability of extracting exactly two winning tickets? (a) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{5}}\) (b) \(\frac{{ }^{9} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{4}}\) (c) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{2}}{{ }^{50} C_{4}}\) (d) \(\frac{{ }^{8} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{3}}\)

Short Answer

Expert verified
The probability is given by option (a): \( \frac{{ }^{8} C_{2} \times{ }^{42} C_{3}}{{ }^{50} C_{5}} \).

Step by step solution

01

Understanding the Problem

The problem involves a situation where 8 winning prizes are randomly distributed in a box of 50 tickets, and a participant draws 5 tickets. We are tasked with finding the probability that the participant draws exactly 2 winning tickets.
02

Calculate the Ways to Pick Winning Tickets

The participant needs to pick exactly 2 winning tickets out of the 8 available. The number of ways to do this is given by the combination formula: \[ ^{8}C_{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]
03

Calculate the Ways to Pick Non-Winning Tickets

Since there are 8 winning tickets, there are 42 non-winning tickets. The participant also needs to draw 3 non-winning tickets (since 5 total tickets are drawn and 2 are winning). The number of ways to do this is given by: \[ ^{42}C_{3} = \frac{42!}{3!(42-3)!} = \frac{42 \times 41 \times 40}{3 \times 2 \times 1} = 11480 \]
04

Calculate the Total Number of Ways to Draw 5 Tickets

The total number of ways to draw any 5 tickets from the 50 is given by the combination formula: \[ ^{50}C_{5} = \frac{50!}{5!(50-5)!} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = 2,118,760 \]
05

Calculate the Probability

The probability of drawing exactly 2 winning tickets is the ratio of the favorable outcomes to the total outcomes, given by:\[ \frac{^{8}C_{2} \times ^{42}C_{3}}{^{50}C_{5}} = \frac{28 \times 11480}{2118760} \] Calculate this to get the probability.
06

Verify and Choose the Correct Option

After calculating the probability, we confirm it matches the given option. Performing the multiplication and division:\[ \frac{28 \times 11480}{2118760} \rightarrow \text{simplifies to option (a).} \]

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Key Concepts

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

Combinatorics
Combinatorics is the field of mathematics that deals with counting, arrangement, and combination of objects. It helps in determining the number of possible outcomes in various scenarios, making it essential for probability problems. When we talk about combinatorics in the context of lotteries or card games, it typically involves finding different ways to select items from a larger set.
  • In our example, combinatorics helps determine how many ways we can select certain numbers of winning and non-winning tickets from a pool of total tickets.
  • This calculation is made possible by using the combination formula, which counts the number of ways to choose a specific number of items from a larger group without consideration for the order.
Understanding and applying combinatorics is crucial for solving probability problems involving a selection process, such as lotteries. It lays the foundation for accurately predicting the likelihood of specific outcomes.
Winning Tickets
When we deal with a lottery scenario, winning tickets become the primary focus of the problem. They are the tickets which fetch a prize, and calculating their probability is a common task in probability exercises.
  • In this exercise, we know there are 8 winning tickets out of a total of 50 tickets.
  • Our task is to calculate the probability of drawing exactly two winning tickets from this box when drawing five tickets in total.
Thinking in terms of winners and non-winners is crucial to break the problem into manageable parts. First, determine how many winning tickets should be drawn, and then figure out how many non-winning tickets accompany them. This helps in structuring the problem for an effective probability calculation.
Probability Calculation
Probability calculation allows us to determine the likelihood of a specific event occurring. It's expressed as a fraction of favorable outcomes over the total possible outcomes. To solve such problems, one must analyze the setup, apply the appropriate formulas, and compute probabilities.
  • The first step involves recognizing how many outcomes are possible in total鈥攃hoosing any 5 tickets from 50 is the total number of possible outcomes, computed using the combination formula.
  • The next step is identifying favorable outcomes, which involve drawing exactly 2 winning and 3 non-winning tickets, each computed separately using combinations.
Ultimately, the probability calculation is completed by dividing the number of favorable outcomes by the number of total outcomes. Probability, thus, becomes a powerful tool in predicting the outcomes of random events like lotteries or card draws.
Combination Formula
The combination formula is a key concept in combinatorics, used to calculate the number of ways to choose a subset of items from a larger set, formally defined as \( ^nC_r \). It's crucial to understand as it allows calculations without considering the order of objects. The general formula is:\[ ^nC_r = \frac{n!}{r!(n-r)!} \]Where \( n \) is the total items and \( r \) is the number of items being chosen from them.
  • Factorials, represented by \( n! \), indicate multiplying a series of descending natural numbers. For instance, \( 5! \) is \( 5 \times 4 \times 3 \times 2 \times 1 \).
  • The formula accounts for all possible unordered selections, which is essential when calculating probabilities within combinations.
In our lottery problem, several combination calculations are made to find the number of ways to draw winning and non-winning tickets, and to determine total possible draws of five tickets. Mastering this formula helps solve complex probability problems efficiently.

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Most popular questions from this chapter

The probability distribution of a random variable \(X\) is given below: \(\begin{array}{llll}X=x_{i} & 2 & 3 & 4 \\ P:\left(X=x_{i}\right) & 1 / 4 & 1 / 8 & 5 / 8\end{array}\) Then its mean is (a) \(27 / 8\) (b) \(5 / 4\) (c) 1 (d) \(4 / 5\)

A die is tossed thrice. A success is getting 1 or 6 on a toss. The mean and the variance of number of successes are: (a) \(\mu=1, \sigma^{2}=2 / 3\) (b) \(\mu=2 / 3, \sigma^{2}=1\) (c) \(\mu=2, \sigma^{2}=2 / 3\) (d) None of theseSolution (a) For binomial distribution, mean \(=n p\) and $$ \begin{aligned} &\text { variance }=n p q \\ &n=3, p=\frac{2}{6}=\frac{1}{3}, q=1-p=1-\frac{1}{3}=\frac{2}{3} \\ &\text { So, } \operatorname{mean}(\mu)=3 \times \frac{1}{3}=1 \\ &\begin{aligned} \text { Variance }\left(\sigma^{2}\right) &=3 \times \frac{1}{3} \times \frac{2}{3} \\ &=\frac{2}{3} \end{aligned} \end{aligned} $$

There are 4 envelopes with addresses and 4 concerning letters. The probability that letter does not go into concerning proper envelope is (a) \(19 / 24\) (b) \(21 / 23\) (c) \(23 / 24\) (d) \(1 / 24\)

A sample of 4 items is drawn at a random without replacement from a lot of 10 items. Containing 3 defective. If \(X\) denotes the number of defective items in the sample then \(P(0

In 324 throws of 4 dice, the expected number of times three sixes occur is (a) 81 (b) 5 (c) 9 (d) 31
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