SHSAT: Probability and counting: choose the right total
Find probabilities from counts, use the opposite event, handle draws without replacement, and count allowed combinations.
How do you decide what goes in the denominator of a probability, and when should you count the opposite event?
Short answer
Count the total equally likely outcomes, then the favorable ones. For "not" or "at least one," find the opposite event's probability and subtract it from 1.
Know first: Simplifying fractions, Multiplying fractions
Step 1 of 5
Probability is favorable over possible
When every outcome is equally likely, the probability of an event is favorable outcomes ÷ total outcomes. Most probability slips come from using the wrong total.
Some events are easier to count backward. The chance something does not happen is 1 minus the chance it does. "At least one" is the opposite of "none," so it is 1 − P(none).
| The question says | Do this |
|---|---|
| not A | 1 − P(A) |
| at least one | 1 − P(none) |
| A and then B, with replacement or separate spins | multiply the probabilities |
| two draws without replacement | multiply, but the second fraction uses the new counts |
| exactly one of two draws | add both orders, such as red-blue and blue-red |
| among the afternoon visitors | use only that group as the total |
| a point chosen in a region | favorable area ÷ total area |
Write the total first. Then count the favorable outcomes inside it.
Step 2 of 5
Update the bag after each draw
Worked example: Two draws without replacement
A box has 6 green marbles and 4 yellow marbles. Two marbles are drawn without replacement. What is the probability that both are green?
- A1/3
- B9/25
- C3/5
- D2/3
- 1
First draw
6 green out of 10: 6/10.
- 2
Second draw
One green marble is gone, so 5 green remain out of 9: 5/9.
- 3
Multiply
6/10 × 5/9 = 30/90 = 1/3.
- 4
See why the other choices are there
9/25 is 6/10 × 6/10, which puts the first marble back. 3/5 stops after one draw. 2/3 is 1 − 1/3, the chance that the two are not both green.
Answer
1/3
Find the wrong step
A fair coin is tossed 4 times. What is the probability of at least one head?
- 1
Each toss gives a head with probability 1/2.
- 2
There are 4 tosses, so add: 1/2 + 1/2 + 1/2 + 1/2 = 2.
What went wrong
A probability can never be greater than 1. Adding counts the outcomes with several heads more than once.
- 3
So a head is certain.
The fix
Use the opposite event. No heads means four tails: (1/2)⁴ = 1/16. So P(at least one head) = 1 − 1/16 = 15/16.
Step 3 of 5
Pick the right total
Check yourself · Question 1
Two fair six-sided number cubes are rolled. What is the probability that the sum is not 7?
Answer: C
There are 6 × 6 = 36 equally likely ordered rolls. Six of them sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). So P(sum is 7) = 6/36 = 1/6, and P(not 7) = 1 − 1/6 = 5/6.
Check yourself · Question 2
A school has 30 juniors and 50 seniors. Of these, 12 juniors and 20 seniors play a sport. Among the seniors, what fraction play a sport?
Answer: B
"Among the seniors" makes the seniors the total: 20 of 50 = 2/5.
Trap answer: 1/4
Using the whole group as the total
- Why it looks right
- 80 is the largest total in the problem, and probability often uses everyone.
- Why it's wrong
- The word among limits the total to the seniors. Students who aren't seniors can't be in the denominator.
- The right answer
- 2/5: 20 sport-playing seniors out of 50 seniors.
Check yourself · Question 3
A bag has 3 red and 2 blue counters. Two are drawn without replacement. What is the probability that exactly one is red?
Answer: C
Red then blue: 3/5 × 2/4 = 6/20. Blue then red: 2/5 × 3/4 = 6/20. Exactly one red can happen either way, so add: 12/20 = 3/5.
Common mistake
I use the same fraction for the second draw.
- Why it's tempting
- The bag looks the same in the problem, and the counts are written only once.
- Do this instead
- Without replacement, the first draw changes the bag. Lower the total by 1, and lower the count of whatever color was drawn.
Step 4 of 5
What to remember
Remember
- Probability = favorable ÷ total. Words like among change the total.
- For not or at least one, subtract the opposite event's probability from 1.
- Without replacement, the second fraction uses the new counts; for exactly one, add both orders.
Step 5 of 5
Practice on a real question
Use what you just learned on this question, then check the explanation.
In your own words
In the practice question, what was your total number of outcomes, and did you count the event directly or use the opposite?