A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random without replacement, what is the probability that both smileys picked are red or both are green?

Introduction / Context:
This question requires combining probabilities of two mutually exclusive events in a single draw without replacement. We have smileys of three different colours in a bag, and we pick two smileys. The event of interest is that either both picked smileys are red or both are green. The two sub events cannot happen at the same time in a two smiley draw, so their probabilities can be added directly.

Given Data / Assumptions:

Red smileys = 5.

Yellow smileys = 6.

Green smileys = 3.

Total smileys = 5 + 6 + 3 = 14.

Two smileys are picked at random without replacement.

We want the probability that both are red or both are green.

Concept / Approach:
We treat this as drawing an unordered pair of smileys from the bag. The total number of possible pairs is 14C2. Favourable outcomes are of two types: both red or both green. We count each type separately (5C2 for red pairs and 3C2 for green pairs), then sum and divide by the total number of pairs. Because a pair cannot be simultaneously all red and all green, the two events are mutually exclusive and we can add their probabilities.

Step-by-Step Solution:
Step 1: Total number of ways to choose 2 smileys from 14 = 14C2.Step 2: Compute 14C2 = (14 * 13) / (2 * 1) = 91.Step 3: Number of ways to choose 2 red smileys from 5 = 5C2 = (5 * 4) / 2 = 10.Step 4: Number of ways to choose 2 green smileys from 3 = 3C2 = (3 * 2) / 2 = 3.Step 5: Total favourable ways (both red or both green) = 10 + 3 = 13.Step 6: Probability = favourable / total = 13 / 91.Step 7: Simplify 13 / 91 by dividing numerator and denominator by 13 to get 1 / 7.

Verification / Alternative check:
We can verify using sequential probabilities. For both red: probability first is red = 5/14, then second is red given first was red = 4/13, so P(both red) = (5/14) * (4/13) = 20 / 182 = 10 / 91. For both green: probability first is green = 3/14, then second green = 2/13, so P(both green) = (3/14) * (2/13) = 6 / 182 = 3 / 91. Summing gives (10 / 91) + (3 / 91) = 13 / 91 = 1 / 7, confirming our combination method.

Why Other Options Are Wrong:
The fraction 3/7 is three times the correct probability and would overstate how often we get two reds or two greens. The fraction 7/91 is half the correct probability and could result from counting only red pairs or only green pairs. The value 0 would be correct only if it were impossible to get two smileys of the same colour, which is not the case. The value 13/91 is equivalent to the unsimplified fraction but is less reduced than 1/7; the simplest correct form is 1/7.

Common Pitfalls:
Students sometimes forget to simplify fractions and may stop at 13/91, not realising that it reduces to 1/7. Others accidentally include red and green mixed pairs in the favourable set, which does not satisfy the condition of both being of the same specified colour. Carefully distinguishing between the two favourable categories and using combinations correctly avoids these errors.

Final Answer:
The probability that both smileys picked are red or both are green is 1/7.