In a game, there are three rooms- I, Il and IIl. Room I contain 2 boxes having gift items and 3 empty boxes, room II contains 3 boxes having gift items and 2 empty boxes, and room III contains 4 boxes having gift items and one empty box respectively. There is an equal probability of each room being chosen by a player. Mr John selects one box from a room chosen at random. The probability that Mr John wins a box having gift items is:
Understanding the Probability Problem
This problem involves calculating the probability of a specific event occurring after a two-stage selection process: first choosing a room, and then choosing a box from that room. We are given information about the contents of each room and the probability of choosing each room.
The key information is:
- There are three rooms: Room I, Room II, and Room III.
- The probability of choosing any room is equal. Since there are 3 rooms, the probability of choosing any specific room is \(\dfrac{1}{3}\).
- Room I contains 2 boxes with gift items and 3 empty boxes. Total boxes in Room I = \(2 + 3 = 5\).
- Room II contains 3 boxes with gift items and 2 empty boxes. Total boxes in Room II = \(3 + 2 = 5\).
- Room III contains 4 boxes with gift items and 1 empty box. Total boxes in Room III = \(4 + 1 = 5\).
We want to find the probability that Mr John selects a box having gift items. This can happen in one of three mutually exclusive ways: (Choose Room I AND get a gift) OR (Choose Room II AND get a gift) OR (Choose Room III AND get a gift).
Calculating Probability of Gift Box from Each Room
First, let's find the probability of getting a gift box given that a specific room is chosen.
- Probability of getting a gift box from Room I (P(Gift | Room I)):
Number of gift boxes in Room I = 2
Total boxes in Room I = 5
P(Gift | Room I) = \(\dfrac{\text{Number of gift boxes}}{\text{Total boxes}} = \dfrac{2}{5}\) - Probability of getting a gift box from Room II (P(Gift | Room II)):
Number of gift boxes in Room II = 3
Total boxes in Room II = 5
P(Gift | Room II) = \(\dfrac{\text{Number of gift boxes}}{\text{Total boxes}} = \dfrac{3}{5}\) - Probability of getting a gift box from Room III (P(Gift | Room III)):
Number of gift boxes in Room III = 4
Total boxes in Room III = 5
P(Gift | Room III) = \(\dfrac{\text{Number of gift boxes}}{\text{Total boxes}} = \dfrac{4}{5}\)
Applying the Law of Total Probability
The problem states that there is an equal probability of choosing each room. Let P(Room I), P(Room II), and P(Room III) be the probabilities of choosing Room I, Room II, and Room III, respectively.
P(Room I) = P(Room II) = P(Room III) = \(\dfrac{1}{3}\)
The total probability of selecting a box with a gift item, P(Gift), is the sum of the probabilities of getting a gift item from each room, weighted by the probability of choosing that room. This is an application of the Law of Total Probability:
P(Gift) = P(Gift | Room I) * P(Room I) + P(Gift | Room II) * P(Room II) + P(Gift | Room III) * P(Room III)
Substitute the calculated probabilities:
\[ P(\text{Gift}) = \left(\dfrac{2}{5} \times \dfrac{1}{3}\right) + \left(\dfrac{3}{5} \times \dfrac{1}{3}\right) + \left(\dfrac{4}{5} \times \dfrac{1}{3}\right) \] \[ P(\text{Gift}) = \dfrac{2}{15} + \dfrac{3}{15} + \dfrac{4}{15} \]Since the denominators are the same, we can add the numerators:
\[ P(\text{Gift}) = \dfrac{2 + 3 + 4}{15} \] \[ P(\text{Gift}) = \dfrac{9}{15} \]Now, simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3:
\[ P(\text{Gift}) = \dfrac{9 \div 3}{15 \div 3} \] \[ P(\text{Gift}) = \dfrac{3}{5} \]Therefore, the probability that Mr John wins a box having gift items is \(\dfrac{3}{5}\).
| Room | Gift Boxes | Empty Boxes | Total Boxes | P(Gift | Room) | P(Room) | P(Gift and Room) |
|---|---|---|---|---|---|---|
| I | 2 | 3 | 5 | \(\dfrac{2}{5}\) | \(\dfrac{1}{3}\) | \(\dfrac{2}{5} \times \dfrac{1}{3} = \dfrac{2}{15}\) |
| II | 3 | 2 | 5 | \(\dfrac{3}{5}\) | \(\dfrac{1}{3}\) | \(\dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{3}{15}\) |
| III | 4 | 1 | 5 | \(\dfrac{4}{5}\) | \(\dfrac{1}{3}\) | \(\dfrac{4}{5} \times \dfrac{1}{3} = \dfrac{4}{15}\) |
Summing the probabilities in the last column gives the total probability of getting a gift:
\[ \text{Total P(Gift)} = \dfrac{2}{15} + \dfrac{3}{15} + \dfrac{4}{15} = \dfrac{9}{15} = \dfrac{3}{5} \]Summary of the Calculation
The steps followed were:
- Identify the probability of choosing each room.
- Calculate the conditional probability of getting a gift box given that a specific room is chosen.
- Use the Law of Total Probability to find the overall probability of getting a gift box.
- Perform the necessary arithmetic and simplify the result.
Revision Table: Probability Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Probability | The measure of the likelihood that an event will occur. | P(Event) = (Number of favorable outcomes) / (Total number of possible outcomes) |
| Conditional Probability | The probability of an event A occurring, given that event B has already occurred. | P(A | B) = P(A and B) / P(B) |
| Law of Total Probability | Used to find the total probability of an event when there are multiple mutually exclusive ways for it to occur. If B1, B2, ..., Bn are mutually exclusive and exhaustive events, then P(A) = \(\sum_{i=1}^{n}\) P(A | Bi)P(Bi). | In this problem: P(Gift) = P(Gift | Room I)P(Room I) + P(Gift | Room II)P(Room II) + P(Gift | Room III)P(Room III) |
| Mutually Exclusive Events | Events that cannot happen at the same time. Choosing Room I and choosing Room II are mutually exclusive events. | P(A and B) = 0 if A and B are mutually exclusive. |
Additional Information: Random Selection and Equal Probability
When a problem states that an item is chosen "at random" from a set of items, it usually implies that each item in the set has an equal probability of being chosen. In this problem, choosing a room "at random" means each of the three rooms has a probability of \(\dfrac{1}{3}\) of being selected. Similarly, when a box is selected from a room, it is implied that each box within that chosen room has an equal chance of being picked.
The structure of this problem is common in probability, combining conditional probability (probability of gift given the room) with the probability of the condition (probability of choosing the room). Understanding how to break down such problems into smaller, manageable steps is crucial for solving them accurately.
Important Questions from Conditional Probability
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
A bike manufacturing factory has two plants P and Q. Plant P manufactures 60 percent of bikes and plant Q manufacture 40 percent. 80 percent of the bikes at plant P and 90 percent of the bikes at plant Q are rated of standard quality. A bike is chosen at random and is found to be of standard quality. What is the probability that it has come from plant P?
20 percent of the pens produced in a factory are of red colour and 4 percent are red and defective. If one pen is picked up at random, then what is the probability of its being defective if it is red?
In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?
