A man draws 3 balls from a jug containing 5 white balls and 7 black balls. He gets Rs. 20 for each white ball and Rs. 10 for each black ball. What is his expectation?
Rs. 42.50
Understanding the Expectation Problem
This problem requires us to calculate the expected monetary gain for a person drawing 3 balls from a jug. The jug contains a mix of white and black balls, and the payout depends on the color of the balls drawn.
Problem Setup
- Total balls in the jug: 5 white balls + 7 black balls = 12 balls.
- Number of balls drawn: 3 balls.
- Reward for each white ball: Rs. 20.
- Reward for each black ball: Rs. 10.
Concept of Expectation
The expectation, or expected value, represents the average outcome of a random event if it were repeated many times. It is calculated by summing the products of each possible outcome's value and its probability. The formula is:
$$E[X] = \sum (\text{Value of Outcome} \times \text{Probability of Outcome})$$We can calculate the expectation for this problem by considering the value obtained from each of the 3 balls drawn separately.
Calculating Expectation Using Linearity of Expectation
A key principle in probability is the linearity of expectation, which states that the expected value of the sum of random variables is equal to the sum of their individual expected values. Let $V$ be the total value received. If $V_1, V_2, V_3$ are the values received from the 1st, 2nd, and 3rd ball drawn, respectively, then $V = V_1 + V_2 + V_3$.
Therefore, the total expectation is $E[V] = E[V_1] + E[V_2] + E[V_3]$.
Expectation for a Single Ball Draw
First, let's determine the probability of drawing each color and the expected value from drawing just one ball.
- The probability of drawing a white ball ($P(W)$) is the number of white balls divided by the total number of balls: $$P(W) = \frac{5}{12}$$
- The probability of drawing a black ball ($P(B)$) is the number of black balls divided by the total number of balls: $$P(B) = \frac{7}{12}$$
The expected value (in Rupees) from drawing a single ball is calculated as:
$$E[\text{Value per ball}] = (P(W) \times \text{Value for white}) + (P(B) \times \text{Value for black})$$ $$E[\text{Value per ball}] = \left(\frac{5}{12} \times 20\right) + \left(\frac{7}{12} \times 10\right)$$ $$E[\text{Value per ball}] = \frac{100}{12} + \frac{70}{12}$$ $$E[\text{Value per ball}] = \frac{170}{12}$$Simplifying this fraction gives:
$$E[\text{Value per ball}] = \frac{85}{6}$$Total Expectation for 3 Balls
Because the expected value for each individual ball drawn remains the same regardless of the order (due to the properties of random sampling without replacement), we can find the total expectation by multiplying the expectation of a single ball draw by the number of balls drawn (3).
$$E[V] = 3 \times E[\text{Value per ball}]$$ $$E[V] = 3 \times \frac{170}{12}$$Now, we simplify the expression:
$$E[V] = \frac{170}{4}$$ $$E[V] = \frac{85}{2}$$ $$E[V] = 42.50$$Final Result
The man's expectation from drawing 3 balls is Rs. 42.50.
Important Questions from Random Variables Basics
The length of time X, needed by an examinee of competition to complete a 1-hour exam, is a random variable with
PDF \(f(x)=\dfrac{6}{5}(x^2+x);0 \le x \le 1.\) , The value of F(0.5) is:If X follows a binomial distribution with n = 6 and \(p=\dfrac{1}{4}\) then the skewness of X is:
If the customers arrive in a shop in Poisson fashion with parameter λ, the fourth raw moment \(\mu_4^{'}\) for the inter-arrival time is:
A discrete random variable X has the probability functions as:
X
0
1
2
3
4
5
6
7
8
f(x)
K
2k
3k
5k
5k
4k
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The value of E(X) is:What percentage of scores falls within three standard deviations from the mean for the normal variate?
