Statistics

In an experiment, a ball is drawn from an urn containing 13 green balls and 7 red balls. If the ball is green, three coins are tossed. Otherwise two coins are tossed. How many elements of the sample space will have a green ball? How many elements of the sample space are there altogether?

Step-by-step solution with explanation

Final Answer

Green-ball elements: . Total sample space: elements.

Step-by-step solution

1

Identify the urn composition and draw

The urn has 13 green and 7 red balls, so 20 total. We draw one ball. The color of the ball determines how many coins we toss next.
2

List outcomes when green ball is drawn

If the ball is green, we toss 3 coins. Each coin has 2 outcomes (H or T), so there are possible coin sequences: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
3

Count sample space elements with green ball

There are 13 distinct green balls, and each can be paired with any of the 8 coin outcomes. So the number of sample space elements involving a green ball is .
4

List outcomes when red ball is drawn

If the ball is red, we toss 2 coins. That gives possible coin sequences: HH, HT, TH, TT.
5

Count sample space elements with red ball

There are 7 distinct red balls, each paired with any of the 4 coin outcomes, giving elements.
6

Find total sample space size

The full sample space is all green-ball outcomes plus all red-ball outcomes. Add them together: .

Understanding this problem

Learning Insight

This is a two-stage experiment, so we use the Fundamental Counting Principle: multiply the choices at each stage. Because the second stage (number of coins) depends on the first stage (ball color), we handle each branch separately and then add the results together. This is the key idea behind tree diagrams and partitioned sample spaces.

Quick Tip

Whenever an experiment branches into separate cases, solve each branch on its own (multiply within the branch), then add all the branches together at the end. Think: multiply ACROSS stages, add ACROSS branches.

Common Mistake

Students often ignore that the balls are distinct and treat the draw as just 'green or red,' computing instead of . Each individual ball creates its own set of outcomes.