Question

A box contains 6 red balls and 4 blue balls. If two balls are drawn at random, then the probability that they are of the same colour is:

• A. \( \frac{7}{15} \)
• B. \( \frac{9}{17} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{4}{15} \)

Hint:

For probability involving selection without replacement: - First find total possible outcomes using combinations. - Then count all favorable cases separately. - If multiple favorable cases exist (like red-red or blue-blue), add them.

Answer 1

The Correct Option is A

Concept: Probability of an event is given by: \[ P(E)=\frac{{Number of favorable outcomes}}{{Total number of possible outcomes}} \] For selecting objects without replacement, combinations are used: \[ {}^nC_r = \frac{n!}{r!(n-r)!} \] Here, two balls can be of the same colour in two ways:

• Both balls are red
• Both balls are blue

We find the probability of each case and add them.
Step 1: Find the total number of ways of drawing 2 balls from 10 balls. Total balls: \[ 6+4=10 \] Number of ways to choose 2 balls from 10: \[ {}^{10}C_2=\frac{10 \times 9}{2}=45 \]
Step 2: Find the number of ways both balls are red. Number of ways of selecting 2 red balls from 6 red balls: \[ {}^6C_2=\frac{6 \times 5}{2}=15 \]
Step 3: Find the number of ways both balls are blue. Number of ways of selecting 2 blue balls from 4 blue balls: \[ {}^4C_2=\frac{4 \times 3}{2}=6 \]
Step 4: Find favorable outcomes for same colour. Favorable outcomes: \[ 15+6=21 \]
Step 5: Calculate the required probability. \[ P({same colour})=\frac{21}{45} \] Simplifying: \[ =\frac{7}{15} \] Therefore, \[ {\frac{7}{15}} \]