1
PROBABILITY / CONDITIONAL PROBABILITY
1.
Events A and B have probabilities P(A) = 0.4, P (B) = 0.65, and P(AB) = 0.85.
(a) Calculate P(AB).
(b) State with a reason whether events A and B are independent.
(c) State with a reason whether events A and B are mutually exclusive.
[6]
(a)
P(A  B) = P(A) + P (B) – P(A  B)
(b)
Because 0.4 × 0.65  0.2 need to see the numbers, not just a statement
Therefore no, not independent
(c)
Because P(A  B)  0
2. Given that P(X) =
(a)
(b)
2
,
3
P(Y|X) =
P(A  B) = 0.2
Not mutually exclusive
2
1
and P(Y|X’) = , find
5
4
P(Y ‘);
P(X′ ∪ Y ′).
[3]
13
2 3 1 3
   =
20
3 5 3 4
(a)
P(Y) =
(b)
P(X  Y) = 1 – P(X  Y)
=1–
3.
4 11
=
15 15
The events A and B are such that P(A) = 0.5, P(B) = 0.3, P(A  B) = 0.6.
(a)
(i)
Find the value of P(A B).
(ii) Hence show that A and B are not independent.
(b)
Find the value of P(B|A).
(b)
Find the value of P(B|A).
(a)
(i)
[6]
Use of P(A B) = P(A) + P(B) - P(A  B)
0.6 = 0.5 + 0.3 - P(A  B)
(ii)
(b)
P(A) P(B) = 0.15

 P(A  B)
 P  B  A =
Use of P B A 
P A
0 .2
= 0.4
0 .5
P(A  B) = 0.2
Hence not independent
2
4. Given that (A∪B)′ = ∅, P(A′|B) =
6
1
and P(A) = , find P(B).
3
7
METHOD 1
c
1
b

 c
bc 3
2
ab 
also
b
3
a  b  c 1  a  b 1
2
6
7
2
1
4
c
(a 
7
7
7 not needed)
P ( B)  b  c 
3
( 0.429)
7
METHOD 2
P( A B) 
P( A  B)
P(B)
P( A  B)  P( A)  1
P (B) 
5.
1
P(B)  P( A  B)
3
1
6
P (B)   1
3
7
3
( 0.429)
7
Given that events A and B are independent with P(A  B) = 0.3 and P(A  B) = 0.3,
find P(A  B).
[ 3]
Method 1: (Venn diagram)
U
A
B
0.3 0.3
P(A  B) = P(A)P(B)
0.3 = 0.6 × P(B)
P(B) = 0.5
Therefore, P(A  B) = 0.8
Method 2:
P(A  B) = P(A) – P(A  B)
0.3 = P(A) – 0.3
P(A) = 0.6
P(A  B) = P(A)P(B) since A, B are independent
0.3 =
0.6 × P(B)
P(B) = 0.5
P(A  B) =
P(A) + P(B) – P(A  B) = 0.6 + 0.5 – 0.3 = 0.8
6.
The independent events A, B are such that P(A) = 0.4 and P(AB) = 0.88. Find
(a)
P(B);
(b)
the probability that either A occurs or B occurs, but not both.
(a)
0.88 = 0.4 + P(B) – 0.4P(B)
0.6P(B) = 0.48  P(B) = 0.8
[6]
3
(b)
METHOD 1
P(A  B) – P(A  B) = 0.88 – 0.32 = 0.56
METHOD 2
P(A)P(B’ ) + P(A’ ) P(B)= 0.4 × 0.2 + 0.6 × 0.8 = 0.56
7. Let A and B be events such that P(A) = 0.6, P(A ∪ B) = 0.8 and P(A | B) = 0.6.
Find P(B).
[6]
EITHER
Using P(A | B) =
P A  B 
P B 
0.6P(B) = P(A  B)
Using P(A  B) = P(A) + P(B) - P(A  B) to obtain 0.8 = 0.6 + P(B) - P(A  B)
Substituting 0.6P(B) = P(A  B) into above equation
or
As P(A | B) = P(A) then A and B are independent events
Using P(A  B) = P(A) + P(B) - P(A) x P(B) to obtain 0.8 = 0.6 + P(B) - 0.6 x P(B)
THEN
8.
0.8 = 0.6 + 0.4 P(B)
P(B) = 0.5
In a school of 88 boys, 32 study economics (E), 28 study history (H) and 39 do not study either
subject. This information is represented in the following Venn diagram.
U (88)
E (32)
H (28)
a
b
c
39
(a)
Calculate the values a, b, c.
(b)
A student is selected at random.
(c)
(i)
Calculate the probability that he studies both economics and history.
(ii)
Given that he studies economics, calculate the probability that he does not study history.
A group of three students is selected at random from the school.
(i)
Calculate the probability that none of these students studies economics.
(ii)
Calculate the probability that at least one of these students studies economics.
[12]
4
U(88)
(a)
E(32)
H(28)
a
b
n (E  H) = a + b + c = 88 – 39 = 49
c
n (E  H) = 32 + 28 – b = 49
60 – 49 = b = 11
39
a = 32 – 11 = 21
c = 28 – 11 = 17
(b)
(c)
11 1

88 8
(i)
P(E  H) =
(ii)
21
21
21
PH ' E  88
P(HE) =
=
(= 0.656) OR Required probability =

32
32
32
PE 
88
(i)
P(none in economics) =
56  55  54
= 0.253
88  87  86
(ii) P(at least one) = 1 – 0.253 = 0.747
 32 56 55   32 31 56  32 31 30
    3      
= 0.747
 88 87 86   88 87 86  88 87 86
OR 
9.
The events B and C are dependent, where C is the event “a student takes Chemistry”, and B is the
event “a student takes Biology”. It is known that
P(C) = 0.4, P(B | C) = 0.6, P(B | C) = 0.5.
(a)
Draw tree diagram.
(b)
Calculate the probability that a student takes Biology.
(c) Given that a student takes Biology, what is the probability that the student takes Chemistry? [4]
(a)
(b) P(B) = 0.4(0.6) + 0.6 (0.5) = 0.24 + 0.30
= 0.54
(c) P(CB) =
P(B  C ) 0.24 4

 (= 0.444, 3 sf)
P(B)
0.54 9
5
10.
Dumisani is a student at IB World College.
7
.
8
The probability that he will be woken by his alarm clock is
1
.
If he is woken by his alarm clock the probability he will be late for school is 4
3
.
5
If he is not woken by his alarm clock the probability he will be late for school is
Let W be the event “Dumisani is woken by his alarm clock”.
Let L be the event “Dumisani is late for school”.
(a)
Draw tree diagram .
(b)
Calculate the probability that Dumisani will be late for school.
(c)
Given that Dumisani is late for school what is the probability that he was woken by his alarm
clock?
[11]
(b)
(a)
Probability that Dumisani will be late
is
(c)
47
7 1 1 3
   =
(0.294)
8 4 8 5 160
P(WL) =
P(W  L)
P(L)
7 1
47

P(L) =
8 4
160
7
35
P(WL) = 32 =
(= 0.745)
47
47
160
P(W  L) =
11.
A painter has 12 tins of paint. Seven tins are red and five tins are yellow. Two tins are chosen at
random. Calculate the probability that both tins are the same colour.
[6]
P(RR) =
7 6 7 
  
12 11  22 
P(YY) =
5 4 5
  
12 11  33 
P (same colour) = P(RR) + P(YY) =
31
(= 0.470, 3 sf)
66
6
12.
(a)
At a building site the probability, P(A), that all materials arrive on time is 0.85. The probability,
P(B), that the building will be completed on time is 0.60. The probability that the materials
arrive on time and that the building is completed on time is 0.55.
(i)
Show that events A and B are not independent.
(ii)
All the materials arrive on time. Find the probability that the building will not be
completed on time.
(b)
There was a team of ten people working on the building, including three electricians and two
plumbers. The architect called a meeting with five of the team, and randomly selected people to
attend. Calculate the probability that exactly two electricians and one plumber were called to
the meeting.
(c) (is that option?) The number of hours a week the people in the team work is normally
distributed with a mean of 42 hours. 10% of the team work 48 hours or more a week.
Find the probability that both plumbers work more than 40 hours in a given week.
[15]
(a) (i) To be independent
(
(
)
)
( )
( )
(
)
( )
( )
Hence A and B are not independent
(ii)
( | )
(
)
( )
(
(b) Probability of 2 electricians and 1 plumber
(c)
X = number of hours worked.
2
X ~ N (42, σ )
P(X  48) = 0.10
P(X < 48) = 0.90
(z) = 0.90
z=
X –

z = 1.28
=>
1.28 =
(z = 1.28155)
48 – 42

=>  = 4.69 (Accept  = 4.68)


P(X > 40) = P  Z 
or
40 – 42 
 = 0.665
4.69 
P(X > 40) = 0.665
Therefore, the probability that one plumber works more than 40 hours per week is 0.665.
The probability that both plumbers work more than 40 hours per week = (0.665)2 = 0.443
)
7
13. There are 30 students in a class, of which 18 are girls and 12 are boys. Four students are selected at random to
form a committee. Calculate the probability that the committee contains
(a)
two girls and two boys;
(b)
students all of the same gender.
[6]
(a) The total number of ways of selecting 4 from 30 = ( )
Number of ways of choosing 2B 2G = ( )( )
(b) Number of ways of choosing 4B = ( ) , choosing 4G = ( )
14.
A team of five students is to be chosen at random to take part in a debate. The team is to be chosen
from a group of eight medical students and three law students. Find the probability that
(a)
only medical students are chosen;
(b)
all three law students are chosen.
[6]
(a) P(only medical students) =
(b) Method 1
P(3 law students)
Method 2
P(3 law students)
16.
Jack and Jill play a game, by throwing a die in turn. If the die shows a 1, 2, 3 or 4, the player who
threw the die wins the game. If the die shows a 5 or 6, the other player has the next throw. Jack plays
first and the game continues until there is a winner. [6]
(a)
Write down the probability that Jack wins on his first throw.
8
(b)
Calculate the probability that Jill wins on her first throw.
(c)
Calculate the probability that Jack wins the game.
(a)
Probability that Jack wins on his first throw =
(b)
Probability that Jill wins on her first throw:
(c)
EITHER
2
(or 0.667)
3
1 2

3 3
=
2
(or 0.222 ).
9
Probability that Jack wins the game:
3
2
1
 2 1 1 2
=
       + ... = 
1
4
3
3 3 3 3
1
9
or
If p is the probability that Jack wins the game then
2
3
2 1 1
p =   p,
so that p = 3 =
1
4
3 3 3
1
9
17. There are 25 disks in a bag. Some of them are black and the rest are white. Two are simultaneously selected at
random. Given that the probability of selecting two disks of the same colour is equal to the probability of selecting
two disks of different colour, how many black disks are there in the bag?
[6]
Let there be n black disks and 25 – n white disks.
Since the two probabilities are equal,
or
P(same colour)
Probabilities are the same, so
P(different colour)
9
18. A new blood test has been shown to be effective in the early detection of a disease. The probability that the
blood test correctly identifies someone with this disease is 0.99, and the probability that the blood test correctly
identifies someone without that disease is 0.95. The incidence of this disease in the general population is 0.0001.
A doctor administered the blood test to a patient and the test result indicated that this patient had the disease.
What is the probability that the patient has the disease?
[6]
Let D be the event that the patient has the disease and S be the event that the new blood test shows that the patient
has the disease. Let D’ be the complement of D, ie the patient does not have the disease.
Now the given probabilities can be written as
P(S|D) = 0.99
P(D) = 0.0001
P(S|D’) = 0.05
Since the blood test shows that the patient has the disease, we are required to find p(D|S).
By Bayes’ theorem,
or
Therefore
19. A girl walks to school every day. If it is not raining, the probability that she is late is
1
.
5
2
1
. The probability that it rains on a particular day is .
3
4
On one particular day the girl is late. Find the probability that it was raining on that day.
If it is raining, the probability that she is late is
[3]
Let P(RL) be the probability that it is raining given that the girl is late.
P(RL) =
P(R  L)
P(L)
P(RL) =
1/ 6
10
=
1 / 6  3 / 20
19
10
20. In a bilingual school there is a class of 21 pupils. In this class, 15 of the pupils speak Spanish as their first
language and 12 of these 15 pupils are Argentine. The other 6 pupils in the class speak English as their first
language and 3 of these 6 pupils are Argentine.
A pupil is selected at random from the class and is found to be Argentine.
Find the probability that the pupil speaks Spanish as his/her first language.
[4]
Let P(S) be the probability that the pupil speaks Spanish.
Let P(A) be the probability that the pupil is Argentine.
Then from the diagram
( | )
or
21.
or
1
. After visiting two shops in
3
succession, he finds he has left his umbrella in one of them. What is the probability that he left his
umbrella in the second shop?
The probability that a man leaves his umbrella in any shop he visits is
[3]
Required probability =
2
9
2 1

9 3

2
.
5
11
22. The probability that it rains during a summer’s day in a certain town is 0.2. In this town, the
probability that the daily maximum temperature exceeds 25°C is 0.3 when it rains and 0.6 when it does not rain.
Given that the maximum daily temperature exceeded 25°C on a particular summer’s day, find the probability that it
rained on that day.
[6]
P(> 25°) = 0.2 × 0.3 + 0.8 × 0.6 = 0.54
P(R >25°) =
23.
1
0.06
= (or 0.111)
9
0.54
Robert travels to work by train every weekday from Monday to Friday. The probability that he catches
the 08.00 train on Monday is 0.66. The probability that he catches the 08.00 train on any other
weekday is 0.75. A weekday is chosen at random.
(a)
(b)
Find the probability that he catches the train on that day.
Given that he catches the 08.00 train on that day, find the probability
that the chosen day is Monday.
(a)
Probability = 0.2 × 0.66 + 0.8 × 0.75 = 0.732
(b)
Probability =
[6]
P(Mon catches train) 0.2  0.66
 11 
=
= 0.180   
P(catchestrain)
0.732
 61 
24. A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from the bag
without replacement. What is the probability that they are of different colours?
[4]
2.
P(different colours) = 1 – [P(GG) + P(RR) + P(WW)]
6
5 
 210  44
 10 9 10 9



 =1– 
=
=1–  
(= 0.677, to 3 sf)
 650  65
 6 25 26 25 26 25 
or
P(different colours) = P(GR) + P(RG) + P(GW) + P(WG) + P(RW) + P(WR)
 10 6   10 10  44
= 4    2   =
(= 0.677, to 3 sf)
 26 25   26 25  65
12
25.
A bag contains 2 red balls, 3 blue balls and 4 green balls. A ball is chosen at random from the bag and
is not replaced. A second ball is chosen. Find the probability of choosing one green ball and one blue
ball in any order.
[4]
R
B
R
3
9
G
R
B
1 1
1
4  4 3
   =  =
3
6 6
8 9 8
OR
B
4
8
4
9
3
9
p(BG or GB) =  
G
3
8
G
R
P(BG or GB) = 2 ×
1
4 3
 =
3
9 8
B
G
26.
Bag A contains 2 red and 3 green balls.
(a)
Two balls are chosen at random from the bag without replacement. Find the probability that 2
red balls are chosen.
Bag B contains 4 red and n green balls.
(b)
Two balls are chosen without replacement from this bag. If the probability that two red balls are
2
chosen is
, show that n = 6.
15
A standard die with six faces is rolled. If a 1 or 6 is obtained, two balls are chosen from bag A,
otherwise two balls are chosen from bag B.
(c)
Calculate the probability that two red balls are chosen.
(d)
Given that two red balls are chosen, find the probability that a 1 or a 6 was obtained on the die.
[13]
(a)
P(RR) =  2  1  = 1
 5  4  10
(b)
P(RR) =
4  3  2
4  n 3  n 15
12 + 7n + n2 = 90
(c)
Forming equation 12 × 5 = 2 (4 + n)(3 + n)
 n2 + 7n – 78 = 0
 n=6
EITHER
P(A) =
1
3
P(B) =
2
3
11
P(RR) = P(A  RR) + P(B  RR) =  1  1    2  2  =
90
 3  10   3  15 
13
OR
1
10
1
3
2
3
(d)
RR
A
B
P(RR) = 1 × 1 + 2 × 2 = 11
3 10 3 15
90
2
15
RR
P(1 or 6) = P(A)
P( A  RR)
P(A|RR) =
=
P( RR)
 
1  
  1 
  3  10  

  
11
90
=
3
11
27. Box A contains 6 red balls and 2 green balls. Box B contains 4 red balls and 3 green balls. A fair cubical die
with faces numbered 1, 2, 3, 4, 5, 6 is thrown. If an even number is obtained, a ball is selected from box A; if an
odd number is obtained, a ball is selected from box B.
(a) Calculate the probability that the ball selected was red.
(b) Given that the ball selected was red, calculate the probability that it came from box B. [6]
(i)
P(R) =
1 6 1 4
  
2 8 2 7
1 6 1 4
  
2 8 2 7
3 2 37
0.661
=  
8 7 56
(a)
P(R) =
(b)
2
16
PB  R 
0.432
PB | R  
 7 =
37
37
P R 
56
28. Bag 1 contains 4 red cubes and 5 blue cubes. Bag 2 contains 7 red cubes and 2 blue cubes.
Two cubes are drawn at random, the first from Bag 1 and the second from Bag 2.
(a) Find the probability that the cubes are of the same colour.
(b) Given that the cubes selected are of different colours, find the probability that
the red cube was selected from Bag 1.
[6]
28
10 38
(a) P(same colour) = P(R  R) + P(B  B) =  4   7    5   2  =      
 9   9   9   9   81   81  81
 0.469
14
(b) P (first red | different) =
P(R and then B)
P(R and then B)  P(B and then R)
 4  2 
  
 9  9 
=
 4  2   5  7 
     
 9  9   9  9 






 4  2   8  
     
 9   9    81   = 8  0.186
43
 38   43  
1     
 81   81  
29. Town has only two bus routes. Route A has twice as many busses as route B. The probability of a bus running
late on route A is
and the probability of it running late on route B is
. At a certain point on the route the
busses run down the same road. If a passenger standing at the bus stop sees a bus running late, use Bayes’ theorem
to find the probability that it is route B bus.
( )
( | )
( )
( | ) ( )
( ) ( | )
( ) ( | )
Piece of cake with tree diagram
30. Mike and Belinda are both keen cyclists and want to see who is the best cyclist . They do this by competing in
series of independent races, and decide that the best cyclist will be the one to win three races. These races only
have two of them as contestants. However, the probability of either of the winning a race is dependent on the
weather. In the rain the probability that Mike will win is 0.8, but when it is dry, the probability that Mike will win
is 0.3. In every race the weather is either defined as rainy or dry. The probability that on the day of the race the
weather is rainy is 0.3.
(a) Find the probability that Mike wins the first race.
(b) Given that Mike wins the first race, what is the probability that the weather is rainy?
(c) Given that Mike wins the first race, what is probability that Mike is the best cyclist?
15
31.
1
. Let X be the number of shots the player
4
takes to score, including the scoring shot. (You can assume that each shot is independent of the
others.)
In a game, the probability of a player scoring with a shot is
(a)
Find P(X = 3).
(b)
Find the probability that the player will have at least three misses before scoring twice.
(c)
Prove that the expected value of X is 4.
(You may use the result (1 – x)–2 = 1 + 2x + 3x2 + 4x3......)
[13]
2
(a)
(b)
1 9
3
P(X = 3) =    
(= 0.141 to 3 sf)
4 64
4
Let the probability of at least three misses before scoring twice = P(3 m)
Let S mean “Score” and M mean “Miss”.
P(3 m) = 1 – [P(0 misses) + P(1 miss) + P(2 misses)]
= 1 – [P(SS) + P(SMS or MSS) + P(MMSS or MSMS or SMMS)]
2
2
2
 1  2
189
1 3 1 3 
= 1 –    2     3     =
(= 0.738 to 3 sf)
256
 4   4   4   4  
 4 
16

(c) E(x) =
xP(x)  1 
for all x
1
1 3
1  32
 2    3   
4
4 4
4  4

  ...


2

 32 
1 
3
1
3



= 1  2   3     ... = 1   (using the given result)
4
4
4
4
4

=
32.
11
 
44
2

1
(4)2 = 4
4
Two women, Ann and Bridget, play a game in which they take it in turns to throw an unbiased sixsided die. The first woman to throw a “6” wins the game. Ann is the first to throw.
(a) Find the probability that
(i)
Bridget wins on her first throw;
(ii)
Ann wins on her second throw;
Ann wins on her nth throw.
(iii)
(b)
(c)
1 25
Let p be the probability that Ann wins the game. Show that p  6  36 p.
Find the probability that Bridget wins the game.
(d)
Suppose that the game is played six times. Find the probability that Ann wins more games than
Bridget.
[17]
(a)
(i)
P(Bridget wins on her first throw)
= P(Ann does not throw a “6”) × P(Bridget throws a “6”) =
(ii)
5 1
5
 =
6 6 36
P(Ann wins on her second throw)
= P(Ann does not throw a “6”) × P(Bridget does not throw a “6”) × P(Ann throws a “6”)
=
(iii)
5 5 1
25
  =
6 6 6 216
P(Ann wins on her nth throw)
= P(neither Ann nor Bridget win on their first (n – 1) throws) ×
5
P(Ann throws a “6” on her nth throw) =  
6
(b)
2 ( n 1)

1
.
6
p = P(Ann wins)
= P(Ann wins on her first throw) + P (both Ann and Bridget do not
win on their first throws) × P(Ann wins from then on)
=
2
1 25
1 5
p
  ×p = 
6 36
6 6
OR
p = P(Ann wins on first throw) + P(Ann wins on second throw)
+ P(Ann wins on third throw) + … .
2
=
4
1 5 1 5 1
 
   
6 6 6 6 6
1
2
4


1 25  1  5   1   5   1 

6 )
+ ... = 
            ... (or
25
6 36 
6
6
6
6
6
       


 1–
36
17
=
1 25

p , as required.
6 36
11
1
6
p  p
36
6
11
Therefore, P(Bridget wins) = 1 – p =
(c)
From part (b),
(d)
P(Ann wins more games than Bridget)
= P(Ann wins 4 games) + P(Ann wins 5 games)
 6  6   5 
4
2
 6  6   5   6 
5
6
5
.
11
+ P(Ann wins 6 games)
64
=               =
(15 × 25 + 36 × 5 + 36) = 0.432.
6
 4  11   11 
 5  11   11   11  11
33. Let A and B be events such that P(A) =
1
1
7
, P(B  A) =
and P(A  B) =
.
5
4
10
(a) Find P(A  B).
(b) Find P(B).
(c) Show that A and B are not independent.
(a) P(A  B) = P(A)  P(B | A) =
1 1  1 

 
5 4  20 
(b) P(A  B) = P(A) + P(B)  P(A  B)  P(B) =
7 1 1
11
 
=
20
10 5 20
METHOD 1
(c)
P(B) =
11
1
and P(B | A) =
20
4
P(B)  P(B | A)  A and B are not independent
METHOD 2
P(A)  P(B) =
1 11 11
1
 
and P(A  B) =
20
5 20 100
P(A  B)  P(A)  P(B)
 A and B are not independent
34. An integer is chosen at random from the first one thousand positive integers.
Find the probability that the integer chosen is
[6]
(a)
a multiple of 4;
(b)
a multiple of both 4 and 6.
(a)
The number of multiples of 4 is 250.
Required probability = 0.25.
(b)
The number of multiples of 4 and 6 is
the number of multiples of 12 = 83.
Required probability = 0.083
18
35.
Only two international airlines fly daily into an airport. UN Air has 70 flights a day and IS Air has 65
flights a day. Passengers flying with UN Air have an 18% probability of losing their luggage and
passengers flying with IS Air have a 23% probability of losing their luggage. You overhear someone
in the airport complain about her luggage being lost.
Find the probability that she travelled with IS Air.
[6]
METHOD 1
Let P(I) be the probability of flying IS Air, P(U) be the
probability flying UN Air and P(L) be the probability of
luggage lost.
P(I | L) =
0.23
=
0.18 
PI  L 
PL 
65
135
70
65
 0.23
135
135
=
299
 0.543 , accept 0.542
551
METHOD 2
Expected number of suitcases lost by UN Air is 0.18 x 70 = 12.6
Expected number of suitcases lost by IS Air is 0.23 x 65 = 14.95
P(I | L) =
36.
14.95
= 0.543
12.6  14.95
A satellite relies on solar cells for its power and will operate provided that at least one of the cells is
working. Cells fail independently of each other, and the probability that an individual cell fails within
one year is 0.8.
(a)
For a satellite with ten solar cells, find the probability that all ten cells fail within one year.
(b)
For a satellite with ten solar cells, find the probability that the satellite is still operating at the
end of one year.
(c)
For a satellite with n solar cells, write down the probability that the satellite is still operating at
the end of one year. Hence, find the smallest number of solar cells required so that the
probability of the satellite still operating at the end of one year is at least 0.95.
[9]
(a)
P(all ten cells fail) = 0.810 = 0.107.
(b)
P(satellite is still operating at the end of one year)
= 1 – P (all ten cells fail within one year) = 1 – 0.107 = 0.893.
P(satellite is still operating at the end of one year) = 1 – 0.8n.
(c)
We require the smallest n for which 1 – 0.8n  0.95. 0.8n  0.05
n
5
   20
4
n
log 20
= 13.4
log1.25
Therefore, 14 solar cells are needed.