Problem 7
Question
In four schools \(B_{1}, B_{2}, B_{3}, B_{4}\) the percentage of girls students is \(12,20,13,17\), respectively. From a school selected at random, one student is picked up at random, and it is found that the student is a girl. The probability that the school selected is \(B_{2}\) is (a) \(\frac{6}{31}\) (b) \(\frac{10}{31}\) (c) \(\frac{13}{62}\) (d) \(\frac{17}{62}\)
Step-by-Step Solution
Verified Answer
The probability that the school selected is \(B_2\) is \(\frac{10}{31}\).
1Step 1: Understand the Information
We have four schools: \(B_1, B_2, B_3, B_4\) with girl percentages \(12\%, 20\%, 13\%,\) and \(17\%\) respectively. A student chosen from a randomly selected school is a girl.
2Step 2: Define the Probabilities
Let \(P(G|B_i)\) be the probability that a student is a girl given that school \(B_i\) is selected. Thus, \[P(G|B_1) = 0.12, \ P(G|B_2) = 0.2, \ P(G|B_3) = 0.13, \ P(G|B_4) = 0.17.\] All schools have an equal chance of being selected, so \(P(B_i) = \frac{1}{4}\) for each \(i\).
3Step 3: Use Bayes' Theorem
To find \(P(B_2|G)\), the probability that the school is \(B_2\) given that a girl has been selected, we apply Bayes' theorem: \[P(B_2|G) = \frac{P(G|B_2) \cdot P(B_2)}{P(G)}.\]
4Step 4: Calculate Total Probability of Selecting a Girl
The total probability of selecting a girl from a random school, \(P(G)\), is calculated as: \[P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) + P(G|B_3)P(B_3) + P(G|B_4)P(B_4) = 0.12\cdot\frac{1}{4} + 0.20\cdot\frac{1}{4} + 0.13\cdot\frac{1}{4} + 0.17\cdot\frac{1}{4}.\] Simplifying gives \[P(G) = 0.155.\]
5Step 5: Calculate the Desired Probability
Substitute values into Bayes' formula: \[P(B_2|G) = \frac{0.2 \cdot \frac{1}{4}}{0.155} = \frac{0.05}{0.155}.\] Simplifying this expression, \[P(B_2|G) = \frac{10}{31}.\]
6Step 6: Verify
Check whether \(\frac{10}{31}\) matches with the given options. The answer is option (b) \(\frac{10}{31}\).
Key Concepts
Understanding ProbabilityBasics of Conditional ProbabilityThe Concept of Random Selection
Understanding Probability
Probability is a way to measure the likelihood of an event happening. It ranges between 0 and 1, where 0 means the event will not happen, and 1 means it will certainly happen. In our example, each school has an equal chance of being selected, indicated by the probability of \( \frac{1}{4} \) for each school. This is because the selection of schools is random.The concept becomes clear by using simple examples:
- If you toss a fair coin, the probability it lands on heads is \(0.5\).
- If a die is rolled, the probability of getting a three is \(\frac{1}{6}\).
Basics of Conditional Probability
Conditional probability takes into account additional information when determining the likelihood of an event. It answers the question, "What's the probability of event A happening given that event B has already occurred?" This is especially useful when outcomes are influenced by prior events.In our exercise, we calculate the probability of selecting a girl from a specific school. Here, the additional information is the school from which the student is selected. This probability is denoted by \(P(G|B_i)\), which means the probability of picking a girl, given a specific school \(B_i\) was selected.To make sense of conditional probability, let's consider:
- If you're given a deck of cards, the probability of pulling an ace is \(\frac{4}{52}\), since there are four aces.
- If you know the card is a spade, the probability of it being an ace, given that it's a spade, is \(\frac{1}{13}\).
The Concept of Random Selection
Random selection is a process where every individual in a population has an equal chance of being chosen. This is key in statistical experiments to ensure fairness and eliminate biases. In this problem, the use of random selection signifies that each school has the same probability of being chosen (\( \frac{1}{4} \)).This fair approach is similar to other scenarios like:
- Randomly selecting a student from a class where every student is equally likely to be selected.
- Drawing a raffle ticket from a bowl where each ticket has the same chance to win.
Other exercises in this chapter
Problem 6
When an ordinary dice is thrown find the probability of getting a number greater than 3 . $$ [\mathrm{MP}-93,97,2002,2004(A)] $$
View solution Problem 6
The probability that at least one of the events \(A\) and \(B\) occurs is \(3 / 5\). If \(A\) and \(B\) occur simultaneously with probability \(1 / 5\), then \(
View solution Problem 7
Two dice are thrown simultaneously. Find the probability of getting a sum 9 in a single throw. $$ \text { [MP-98, 2003, 2004 (C)] } $$
View solution Problem 7
From a pack of 52 cards 2 cards are drawn in succession one by one without replacement. The probability that both are aces is (a) \(2 / 13\) (b) \(1 / 51\) (c)
View solution