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
- Match the LIST-I with LIST-II
LIST-I LIST-II A. Sampling error I. Values that describe the characteristics of a population B. Parameters II. The standard deviation of the sampling distribution C. Sampling fraction III. Variation among sample statistics due to change D. Standard error IV. Proportion of the population contained in a sample
Choose the correct answer from the options given below: Given below are two statements one is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A): Additive model considers the total of various components resulting in the given
values of the over all time series and it is stated as:
$Y=T+C+S+I$
Reason (R): Multiplicative model assumes that the various components interact in a
multiplicative manner to produce the given values of the over all time series and
can be stated as $Y=T \times C \times S \times I$
In the light of the above statements, choose the most appropriate answer from the options given
below:
Match List - I with List - II.
List - I List - II A. Sample proportion P = x/n I. $\frac{\sigma}{\sqrt{n}}$ B. Sample variance $S_1^2 = \frac{1}{n-1} \sum_{i=1}^{n}(xi-x̄)^2$ II. $\sigma^2 \sqrt{\frac{2}{n}}$ C. Sample variance $S^2 = \frac{1}{n} \sum_{i=1}^{n}(xi-x̄)^2$ III. $\sigma^2 \sqrt{\frac{2}{(n-1)}}$ D. Sample mean $\bar{x} = \frac{1}{n} \sum_{i=1}^{n}xi$ IV. $\sqrt{\frac{\pi(1-\pi)}{n}}$
Choose the correct answer from the options given below :- The measurement or determinations of length, height, weight, volume, area, pressure etc. comes under:
- Which of the following options explain what a 'mean' assumes ?
(i) an interval level of measurement
(ii) a nominal level of measurement
(iii) an ordinal level of measurement
(iv) a ratio level of measurement
