Question

There are 100 tickets in a box numbered \(00,01, \ldots \ldots .99\). One ticket is drawn at random. If \(A\) is the event that sum of the digits of the number is 7 and \(B\) is the event that product of digit is 0 . Then \(\mathrm{P}(\mathrm{A} / \mathrm{B})=\) (a) \((2 / 13)\) (b) \((2 / 19)\) (c) \((1 / 50)\) (d) None

Short answer

The short answer is: The probability of P(A/B) is \(\frac{2}{19}\). The correct option is (b) \(\frac{2}{19}\).

Step-by-step solution

Finding the Sample Space for Event B

First, let's find the sample space for event B where the product of digits is 0. This happens when either of the digits of a number is 0. List out all the numbers that have a digit as 0, keeping in mind that the ticket numbers are from 00 to 99. Sample space for Event B = {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40, 50, 60, 70, 80, 90} There are a total of 19 elements in the sample space for event B.

Finding the Sample Space for Event A and B

Now, let's find the sample space for numbers that belong to both event A (sum of digits equals 7) and event B (product of digits equals 0). To do this, we look at the numbers whose digits sum up to 7 and has at least one digit as 0. Sample space for Event A and B = {07, 70} There are 2 elements in the sample space for event A and B.

Calculating the Conditional Probability

Now that we have found the sample space of Event B and Event A and B, we can find the conditional probability of P(A/B) as follows: P(A/B) = (Number of elements in the sample space of A and B) ÷ (Number of elements in the sample space of B) P(A/B) = \(\frac{2}{19}\) So, the probability of P(A/B) is \(\frac{2}{19}\). The correct answer is (b) \(\frac{2}{19}\).

Key concepts

Probability

Probability is a concept that measures the likelihood of an event occurring. It's a fundamental part of statistics and is used to predict the chances of various outcomes based on certain conditions.

Mathematically, probability is expressed as a number between 0 and 1. An event with a probability of 0 means it is impossible for the event to occur, while a probability of 1 means the event is certain to happen. Events with probabilities closer to 1 are more likely to occur than those with probabilities closer to 0.

To calculate the probability of a single event, you would use the following formula:
\[ P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \]
Using the exercise example, we can see that the process to find the conditional probability requires understanding of how likely one event is to occur given that another event has occurred.

Sample Space

The sample space of an experiment is the set of all possible outcomes of that experiment. It is often represented by the symbol \( S \), and each specific outcome is called an 'element' or a 'sample point'.

For instance, if you toss a six-sided die, the sample space \( S \) is \( {1, 2, 3, 4, 5, 6} \), since these are all the possible outcomes. In the context of the textbook exercise, the sample space was defined as all possible two-digit numbers that could be drawn from a box, with some numbers fulfilling specific conditions.

When calculating probabilities, a well-defined sample space is crucial for determining the total number of possible outcomes, which is a fundamental part of many probability calculations. For example, in the exercise, identifying the sample space of ticket numbers where the product of digits is zero (Event B) was an essential first step.

Event Intersection

Event intersection refers to a situation where two events both occur at the same time. In probability, this concept is used to calculate the likelihood of both events happening simultaneously.

In probability notation, the intersection of two events, A and B, is denoted as \( A \cap B \), and it represents the set of outcomes that are part of both events A and B. For a simple example, if A is the event 'roll an even number on a die' and B is the event 'roll a number greater than three', then \( A \cap B \) will be the outcome set \( {4, 6} \), since 4 and 6 are both even and greater than three.

In the exercise, the intersection was between Event A (sum of the digits equals 7) and Event B (product of digits equals 0). Finding the intersection of these two events was necessary to determine the conditional probability \( P(A/B) \), effectively zeroing in on the overlap between these two specific conditions within the broader sample space of ticket possibilities.