Probability review Problem set A One question at a time

A family has two children.

1. (a) Given that at least one is a girl, find the probability that both are girls.

   • Sample space (older first): BB, BG, GB, GG
   • At least one girl rules out BB. Left: BG, GB, GG
   • Only GG has two girls: \(P=\frac{1}{3}\)

2. (b) Given that the younger child is a girl, find the probability that both are girls.

   • Younger is a girl: the outcomes ending in G are BG and GG. Left: BG, GG
   • \(P=\frac{1}{2}\)

A broadband connection fails on any given day with probability 0.08, independently of other days.

1. (a) Find the probability that the connection fails on at least one of 3 days.

   • \(P(\text{none})=0.92^{3}=0.779\)
   • \(P(\text{at least one})=1-0.779=0.221\)

2. (b) Find the probability that the connection fails on all of 3 days.

   • Multiply: \(0.08^{3}=0.000512\)

A drawer holds 9 black socks and 6 white socks. Two socks are taken out at random one after the other, without replacement.

1. (a) Given that the first is black, find the probability that the second is also black.

   • After one black is taken: 8 black left out of 14.
   • \(P=\frac{8}{14}=\frac{4}{7}\)

2. (b) Find the probability that both are black.

   • \(P=\frac{9}{15}\times\frac{8}{14}=\frac{12}{35}\)

3. (c) Is it more likely that the two are the same colour or different colours? Justify your answer.

   • Both black: \(\frac{12}{35}\). Both white: \(\frac{6}{15}\times\frac{5}{14}=\frac{1}{7}\).
   • Same: \(\frac{17}{35}\). Different: \(1-\frac{17}{35}=\frac{18}{35}\)
   • \(\frac{18}{35}\) is larger, so different is more likely.

Of 80 members of an arts club, 30 paint, 34 sculpt and 26 print. 10 paint and sculpt, 9 sculpt and print, 8 paint and print, and 5 do all three.

1. (a) Draw a Venn diagram and fill in every region, including the region outside the circles.

   • Start in the middle: 5 in all three.
   • Two only: P and S only: 10 − 5 = 5. S and R only: 9 − 5 = 4. P and R only: 8 − 5 = 3.
   • One only: P only 17, S only 20, R only 14.
   • Outside: 80 − 68 = 12

2. (b) Find the probability that a person chosen at random prints but does not paint or sculpt.

   • \(P=\frac{14}{80}=\frac{7}{40}\)

3. (c) Given that a person chosen at random prints, find the probability that they also sculpt.

   • Circle R holds 26 people. 9 of them are in S too.
   • \(P=\frac{9}{26}\)

Each of 32 pupils has at least one of a phone (P), a smart watch (W) or a tablet (T). 15 have a phone, 16 have a smart watch and 14 have a tablet. 5 have a phone and a smart watch, 6 have a smart watch and a tablet, and 4 have a phone and a tablet.

1. (a) Let x be the number of pupils who have all three items. Draw a Venn diagram showing this information, with each region in terms of x.

   • All three: x. Two only: take x from each overlap: 5 − x, 6 − x, 4 − x.
   • One only: circle total, minus the two overlaps, plus x (taken off twice). P only: 15 − 5 − 4 + x = 6 + x.
   • W only: 5 + x. T only: 4 + x.

2. (b) Show that \(x+30=32\).

   • Add all seven regions. The x terms give \(1-3+3=1\) lots of x.
   • The numbers add to 30, so the total is \(x+30\).
   • Everyone has at least one item, so \(x+30=32\).

3. (c) Find the probability that a pupil chosen at random has all three.

   • \(x=32-30=2\)
   • \(P=\frac{2}{32}=\frac{1}{16}\)

4. (d) Given that a pupil does not have a tablet, find the probability that they have a phone.

   • Outside circle T: P only 8, W only 7, P and W only 3, total 18.
   • Of these, 11 are in circle P: \(P=\frac{11}{18}\)

5. (e) Given that a pupil has exactly two of the three, find the probability that they have a smart watch and a tablet.

   • Exactly two: 3 + 4 + 2 = 9.
   • \(P=\frac{4}{9}\)

A family has three children. Write B for a boy and G for a girl, eldest first.

1. (a) List the 8 equally likely outcomes.

   • Write each outcome in order: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

2. (b) Given that at least one is a boy, find the probability that all three are boys.

   • At least one is a boy rules out only the outcome with none of that kind. 7 outcomes are left.
   • Exactly one of them has all three the same kind.
   • \(P=\frac{1}{7}\)

3. (c) Given that the eldest child is a boy, find the probability that all three are boys.

   • The eldest child is a boy: the 4 outcomes starting with that kind are left.
   • One of those 4 has all three the same kind.
   • \(P=\frac{1}{4}\)

On a school day, Ben is late with probability 0.1 and Cara is late with probability 0.2. They are late independently. Find the probability that at least one of them is late.

• Independent, so \(P(\text{neither})=(1-0.1)(1-0.2)=0.72\)
• \(P(\text{at least one})=1-0.72=0.28\)