AI SL · 13.1
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Probability review: the boy or girl problem
13.1.9
A mother has two children. One of them is a boy. Find the probability that both are boys.
Have a go first. Then open the working one line at a time.
BB, BG, GB, GG
Older child first. Each outcome has probability \(\frac{1}{4}\).
Cross out GG
At least one boy: only GG is ruled out.
Left: BB, BG, GB. Only BB has two boys.
Three equally likely outcomes remain.
\(P=\frac{1}{3}\)
Same as \(\frac{P(BB)}{P(\text{at least one boy})}=\frac{1/4}{3/4}\).
A mother has two children. The older one is a boy. Find the probability that both are boys.
Have a go first. Then open the working one line at a time.
BB, BG, GB, GG
Older child first. Each outcome has probability \(\frac{1}{4}\).
Cross out GB and GG
The older child is a boy: any outcome starting with G goes.
Left: BB, BG
Two equally likely outcomes remain.
\(P=\frac{1}{2}\)
The condition names a child, so the younger child is free to be a boy or a girl. This is why the answers differ.
A tin holds 7 plain biscuits and 5 chocolate biscuits. Two are taken out at random, one after the other, without replacement. Is it more likely that the two biscuits are the same type or different types?
Have a go first. Then open the working one line at a time.
\(\frac{7}{12}\times\frac{6}{11}=\frac{42}{132}\)
After one plain biscuit, 6 plain and 11 in all.
\(\frac{5}{12}\times\frac{4}{11}=\frac{20}{132}\)
After one chocolate, 4 chocolate and 11 in all.
\(P=\frac{62}{132}=\frac{31}{66}\)
Add the two branches.
\(P=1-\frac{31}{66}=\frac{35}{66}\)
0.530 is larger than 0.470: different is more likely.
A train is late on any given day with probability 0.15, independently of other days. Find the probability that it is late on at least one of 4 days.
Have a go first. Then open the working one line at a time.
\(P(\text{at least one})=1-P(\text{none})\)
'At least one' has many cases. 'None' has one.
\(0.85^4=0.522\)
Not late has probability 1 − 0.15 = 0.85. Four independent days: multiply.
\(1-0.522=0.478\)
About 0.478.
In a group of 60 pupils, 28 play football (F), 25 play hockey (H) and 22 play rugby (R). 9 play football and hockey, 8 play hockey and rugby, 7 play football and rugby, and 4 play all three. (a) Draw a Venn diagram. (b) Find the probability that a pupil plays exactly one of the three sports. (c) Given that a pupil plays rugby, find the probability that they also play hockey.
Have a go first. Then open the working one line at a time.
4
Start in the middle.
F and H only: 9 − 4 = 5. H and R only: 8 − 4 = 4. F and R only: 7 − 4 = 3.
Take the middle out of each overlap.
F only 16, H only 12, R only 11. Outside: 60 − 55 = 5.
For F only: 28 − 5 − 3 − 4 = 16.
\(P=\frac{16+12+11}{60}=\frac{39}{60}=\frac{13}{20}\)
Add the three single-circle regions.
\(P=\frac{8}{22}=\frac{4}{11}\)
Circle R holds 22 pupils. 8 of them are also in H.
Each of 34 pupils has at least one of a cat (C), a dog (D) or a fish (F). 20 have a cat, 16 have a dog and 16 have a fish. 7 have a cat and a dog, 6 have a dog and a fish, and 8 have a cat and a fish. Let x be the number who have all three. (a) Show that \(x+31=34\). (b) Find x.
Have a go first. Then open the working one line at a time.
C and D only: 7 − x. D and F only: 6 − x. C and F only: 8 − x.
The overlaps include the x pupils in the middle.
C only: 20 − (7 − x) − (8 − x) − x = 5 + x. D only: 3 + x. F only: 2 + x.
Take away the two overlaps and the middle, so x comes back once.
\(x+(7-x)+(6-x)+(8-x)+(5+x)+(3+x)+(2+x)=x+31\)
The x terms: 1 − 3 + 3 = 1. Everyone has at least one, so the total is 34. This shows \(x+31=34\).
\(x=34-31=3\)
Check: every region is at least 0 when x = 3.