Probability Calculation: Sequential Fruit Removal
This problem involves calculating the probability of two dependent events occurring in sequence.
Initial Setup
- Total fruits in the box: 3 apples + 2 oranges = 5 fruits.
- We are removing two fruits randomly in succession (without replacement).
- We need the probability that the first fruit is an apple AND the second fruit is an orange.
Step 1: Probability of the First Fruit Being an Apple
The probability of the first fruit selected being an apple is the number of apples divided by the total number of fruits.
Let $P(\text{1st is Apple})$ denote this probability.
$ P(\text{1st is Apple}) = \frac{\text{Number of Apples}}{\text{Total Fruits}} = \frac{3}{5} $
Step 2: Probability of the Second Fruit Being an Orange (Given the First was an Apple)
After removing one apple, the box composition changes:
- Remaining apples: 3 - 1 = 2
- Remaining oranges: 2
- Total remaining fruits: 5 - 1 = 4
The probability of the second fruit selected being an orange, given that the first was an apple, is the number of remaining oranges divided by the total remaining fruits.
Let $P(\text{2nd is Orange} | \text{1st is Apple})$ denote this conditional probability.
$ P(\text{2nd is Orange} | \text{1st is Apple}) = \frac{\text{Number of Oranges}}{\text{Total Remaining Fruits}} = \frac{2}{4} = \frac{1}{2} $
Step 3: Combined Probability
The probability of both events happening in sequence is the product of their individual probabilities (using the rule $P(A \text{ and } B) = P(A) \times P(B|A)$).
$ P(\text{1st is Apple AND 2nd is Orange}) = P(\text{1st is Apple}) \times P(\text{2nd is Orange} | \text{1st is Apple}) $
$ P = \frac{3}{5} \times \frac{2}{4} $
$ P = \frac{6}{20} $
Simplifying the fraction:
$ P = \frac{3}{10} $
Conclusion
The probability that the first fruit removed is an apple and the second is an orange is $\frac{3}{10}$.
Important Questions from Conditional Probability
A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability 𝑃(ℎ𝑒𝑎𝑑) = 0.5 and for a fake coin, 𝑃(ℎ𝑒𝑎𝑑) = 1. You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)
- A clinic specializes in testing for a disease D. The result of the test can be either positive or negative.
A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.
If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places) - Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let $E_1$ be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let $E_2$ be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.
The conditional probability P($E_1$ | $E_2$) is equal to ____________. (rounded off to one decimal place) - An experiment is performed by rolling an unbiased die once. Events A and B are defined as follows:
Event A: The outcome of the rolled die is a prime number.
Event B: The outcome of the rolled die is an odd number less than 4.
The conditional probability of A given B, denoted by P(A|B), is ______ (rounded off to one decimal place). - Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is ______ (rounded off to two decimal places).
