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).
Understanding the Problem
This question asks for a conditional probability. We need to find the probability that a randomly selected ball came from Bag I, given that the selected ball is black. We can use Bayes' Theorem.
Defining Events and Probabilities
Let B1 be the event that Bag I is chosen, and B2 be the event that Bag II is chosen. Let B be the event that a black ball is drawn.
- Probability of choosing Bag I: $P(B1) = 1/2$
- Probability of choosing Bag II: $P(B2) = 1/2$
- Contents of Bag I: 4 white, 6 black balls (Total = 10).
- Contents of Bag II: 4 white, 3 black balls (Total = 7).
- Probability of drawing a black ball given Bag I was chosen: $P(B | B1) = 6 / 10$
- Probability of drawing a black ball given Bag II was chosen: $P(B | B2) = 3 / 7$
Calculating Total Probability of Drawing a Black Ball
We first find the overall probability of drawing a black ball, $P(B)$, using the law of total probability:
$P(B) = P(B | B1) * P(B1) + P(B | B2) * P(B2)$
Substituting the values:
$P(B) = (6 / 10) * (1 / 2) + (3 / 7) * (1 / 2)$
$P(B) = 3 / 10 + 3 / 14$
To add these fractions, find a common denominator (70):
$P(B) = (3 * 7) / 70 + (3 * 5) / 70$
$P(B) = 21 / 70 + 15 / 70$
$P(B) = 36 / 70$
Simplifying the fraction:
$P(B) = 18 / 35$
Applying Bayes' Theorem
Now we apply Bayes' Theorem to find the probability that the black ball was drawn from Bag I given that it is black, $P(B1 | B)$:
$P(B1 | B) = [ P(B | B1) * P(B1) ] / P(B)$
Substitute the calculated and given probabilities:
$P(B1 | B) = [ (6 / 10) * (1 / 2) ] / (18 / 35)$
$P(B1 | B) = (3 / 10) / (18 / 35)$
To divide by a fraction, multiply by its reciprocal:
$P(B1 | B) = (3 / 10) * (35 / 18)$
$P(B1 | B) = (3 * 35) / (10 * 18)$
$P(B1 | B) = 105 / 180$
Simplify the fraction by dividing numerator and denominator by their greatest common divisor (15):
$P(B1 | B) = 7 / 12$
Final Result and Rounding
Convert the fraction to a decimal:
$7 / 12 ≈ 0.58333...$
Rounding to two decimal places as required:
$P(B1 | B) ≈ 0.58$
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) - A box contains three apples and two oranges. Two fruits are removed randomly in succession. The probability that the first is an apple and the second is an orange is _____ .
- 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).
