AI SL · 13.1
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Conditional probability
13.1.6
Fill the tree from a bag of counters, with or without replacement, or type your own probabilities in the orange boxes. The grey box on each pair is 1 minus its partner.
The four end products
Choose a bag or type your own probabilities.
Type P(A), P(B) and P(A ∩ B). The four regions fill in. Choose what to shade and read the working. The circles are not drawn to scale.
Type three probabilities to fill the diagram.
A bag contains 4 blue and 7 yellow counters. Two counters are taken out at random, one after the other, without replacement. Given that the first counter is blue, find the probability that the second counter is yellow.
Have a go first. Then open the working one line at a time.
4 blue + 7 yellow = 11. After a blue: 3 blue + 7 yellow = 10
Without replacement the counter does not go back. The bag now has 10 counters.
7 of the 10 counters are yellow
On the tree this is the branch that starts at 'Blue': second-stage probabilities are conditional.
\(P(Y2 | B1)=\frac{7}{10}=0.7\)
The two branches from Blue add to 1: 3/10 + 7/10.
Two cards are taken at random from a standard pack of 52 cards, without replacement. The first card is a queen. Find the probability that the second card is also a queen.
Have a go first. Then open the working one line at a time.
52 − 1 = 51 cards; 4 − 1 = 3 queens
Take the first card away from the total and from its own kind.
\(P=\frac{3}{51}=\frac{1}{17}\)
Write the fraction in lowest terms.
The table shows the options chosen by 80 students. A student is chosen at random. Given that the student is a boy, find the probability that the student chose music.
Have a go first. Then open the working one line at a time.
Girls 40, Boys 40, total 80
Add the columns.
\(P(\text{Music}\mid\text{Boy})=\frac{P(\text{Music and Boy})}{P(\text{Boy})}=\frac{6/80}{40/80}\)
Both probabilities share the denominator 80.
\(P=\frac{6}{40}=\frac{3}{20}=0.15\)
The same as keeping the Boys column and dividing by its total.
Of 40 students, 24 study music, 15 study drama and 9 study both. A student is chosen at random. Given that the student studies music, find the probability that the student studies drama.
Have a go first. Then open the working one line at a time.
Both: 9
Always fill the overlap first.
Music only: 24 − 9 = 15. Drama only: 15 − 9 = 6.
The 24 includes the 9 who study both.
40 − (15 + 9 + 6) = 10
Everyone else goes outside the circles.
Music: 15 + 9 = 24. Of these, drama: 9.
Ignore everything outside the music circle.
\(P(D | M)=\frac{9}{24}=\frac{3}{8}=0.375\)
The denominator is 24, not 40.
A student is chosen at random. The probability that the student cycles to school but does not swim is 0.30. The probability that the student cycles and swims is 0.20. The probability that the student does neither is 0.25. Given that the student cycles, find the probability that the student swims.
Have a go first. Then open the working one line at a time.
Swims only: 1 − (0.30 + 0.20 + 0.25) = 0.25
All four regions add to 1.
P(cycles) = 0.30 + 0.20 = 0.50
This is the new total, in place of 1.
\(P(S\mid C)=\frac{P(S\cap C)}{P(C)}=\frac{0.20}{0.50}=0.4\)
This is the definition, and it is the same as the shaded part of the circle.