Understanding the Probability Problem
The problem asks for the probability of selecting two balls of the same colour when drawing two balls at random from a bag. The bag contains a specific number of black and white balls.
Key Information:
- Number of black balls = 5
- Number of white balls = 4
- Total number of balls = 5 + 4 = 9
- Number of balls selected = 2
- Selection is done at random.
Calculating Total Possible Outcomes
We need to find the total number of ways to select any 2 balls from the 9 balls available. Since the order of selection doesn't matter, we use combinations. The formula for combinations is \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items, and \(k\) is the number of items to choose.
Total ways to select 2 balls from 9 is:
\(C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36\)So, there are 36 possible pairs of balls that can be selected.
Calculating Favourable Outcomes
The condition is that both selected balls must be of the same colour. This can happen in two mutually exclusive ways:
- Both balls selected are black.
- Both balls selected are white.
Scenario 1: Both Balls are Black
Number of ways to select 2 black balls from the 5 black balls available:
\(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10\)Scenario 2: Both Balls are White
Number of ways to select 2 white balls from the 4 white balls available:
\(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6\)Determining the Probability
The total number of favourable outcomes is the sum of the outcomes from the two scenarios (both black OR both white):
Total Favourable Outcomes = (Ways to select 2 black balls) + (Ways to select 2 white balls)
\(\text{Total Favourable Outcomes} = 10 + 6 = 16\)The probability of an event is calculated as:
\(P(\text{Event}) = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}\)Therefore, the probability of selecting two balls of the same colour is:
\(P(\text{Same Colour}) = \frac{16}{36}\)Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:
\(P(\text{Same Colour}) = \frac{16 \div 4}{36 \div 4} = \frac{4}{9}\)Similar Questions
- What is the value of \(n\) ?
- What is the mean of the distribution ?
- What is the value of \(p + q\) ?
- What is \(\sum_{i=1}^n x_i f_i\) equal to ?
- What is the mean deviation of the largest five observations ?
- What is the variance of the largest five observations ?
What is the median height of the class?
- If the mean of 20 observations, namely \(x_1, x_2, x_3, \dots, x_{20}\) is 1.414, then what is the value of \(\sum_{i=1}^{20} 100(2x_i + 4)\) ?
What is the total number of students whose height is less than or equal to 165 cm?
The height which occurs most frequently in the class is
Important Questions from Statistics
Methods of presentation of data are:
Consider 35 i.i.d. observations X1, X2, ...., X15 and Y1, Y2, ..., Y20. Let R be the Wilcoxon's rank sum statistic based on the ranks of the X' s in the combined sample. Then the expected value of R is
Let I, J > 5. Consider two-way ANOVA, where the observations satisfy the linear model
yij = α + βi + γj + εij, 1 ≤ i ≤ I, 1 ≤ j ≤ J
E(εij) = 0, Var(εij) = σ2, \(\sum_{i=1}^{!} β_i=\sum_{j=1}^j γ_j=0 .\)
In this set - up
Let X1,X2,...,Xn be a random sample from exponential distribution with mean θ. Which of the following is NOT a sufficient statistic for θ?
A newly developed algorithm for random number generation needs to be tested. The first step is to check whether the sequence of numbers generated can be considered a random sample from the uniform distribution on the interval (0, 1). Which of the following is an appropriate nonparametric test?
