SOLUTIONS
QUESTION ONE A
Given that P and Q are events, when is P and Q?
Independent;
Events P and Q are independent if the occurrence of one event does not affect the occurrence of
another event. For instance if P is occurring event Q will also be occurring.
Mutually exclusive
P and Q are said to be mutually exclusive events when the occurrence one event affects the
occurrence to occur, for instance event P and Q cannot occur at the same time e.g. one cannot
live and die at the same time.
Exhaustive events
Events P and Q are said to be exhaustive if at least one of the events must occur. Events are said
to be exhaustive if regardless of the circumstance one, event must always occur for instance
during a dice role, its either event 1, 2, 3, 4, 5 and 6 will occur.
QUESTION ONE B
Machine A = 20% OF total production
Machine B = 80%
Let event A and B be the respective events for machines A and B contribution to total production
Machine A = 5%
Machine B = 10%, Defective
Let x be defectiveness
=𝑃(𝐴) = 0.2
=𝑃(𝐵) = 0.8
=𝑃(𝑥𝐴 ) = 0.05
=𝑃(𝑥𝐵 ) = 0.1
Hence the probability that the randomly selected item was from machine A is𝑃(𝑥𝐴/𝑥𝐴 ), Bayer’s
theorem
𝑃(𝑥𝐴/𝑥𝐴 ) =
=
𝑃(𝐴) ∗ 𝑃(𝑥𝐴 )
𝑃(𝐴)𝑃(𝑥𝐴 ) + 𝑃(𝐵) ∗ 𝑃(𝑥𝐴 )
0.2 ∗ 0.05
(0.2 ∗ 0.05) + (0.8 ∗ 0.1)
= 0.1111
𝑷(𝒙𝑨/𝒙𝑨 ) = 𝟏𝟏. 𝟏𝟏%
QUESTION ONE C
a)
Probability that the selected stock is from AMEX and unchanged
25
500
= 𝟎. 𝟎𝟓
b) Marginal probability that the selected stock is from NYSE.
Marginal probability is the probability of one event in the presence of other random variables.
Therefore the marginal probability that stocks is unchanged is from NYSE is as follows.
Since AMEX has 200 stock and of the 200 is one unchanged probability is
300
500
= 𝟎. 𝟔
c)
Conditional probability is the probability of one event given the occurrence of another event.
Therefore, the probability that stock is unchanged given it is from AMEX is as follows
25
200
= 0.125
QUESTION TWO
A) I
All members line up
8𝑃8
8! = 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1
= 𝟒𝟎, 𝟑𝟐𝟎
A) II
Permutation will be required to be used in this case (ways of choosing President, Vice President
and secretary.
𝑃𝑛𝑁 =
𝑁!
(𝑁 − 𝑛)!
Where N=8 and n= 5
𝑃𝑛𝑁 =
8!
(8 − 3)! 3!
𝑷𝑵
𝒏 = 𝟑𝟑𝟔
A) III
Ways to choose 5 members does not require order as such to find the number of ways of
choosing five members to attend a normal general meeting we use combinations
𝐶𝑛 =
=
𝑁!
𝑛! (𝑁 − 𝑛)!
8!
5! (8 − 5)
𝑪𝒏 = 𝟓𝟔
QUESTION 2 B
I)
𝐵(𝑥: 𝑛, 𝑝) = 𝑛𝐶𝑥 ∗ 𝑃 𝑥 ∗ (1 − 𝑝)𝑛−𝑥
Where;
n= total number of events
? =total number of successful events
P= probability of success
1-p probability of failure
𝑛!
∗ (𝑝)𝑥 ∗ 𝑞 𝑛−𝑥
(𝑛 − 𝑥)𝑥!
𝑃(𝑇𝑤𝑜 𝐹𝑎𝑣𝑜𝑢𝑟 𝑇ℎ𝑒 𝐵𝑎𝑛) = 5𝐶2 ∗ 0.62 ∗ (1 − 0.6)5−2
5!
∗ 0.62 (1 − 0.6)^3
(5 − 2)2!
10 ∗ 0.36(0.064)
𝟎. 𝟐𝟑𝟎𝟒
II)
P(Less than four favour the ban) = P (0) + P (1) + P (2) + P (3)
𝑃(0) = 5𝐶0 ∗ 0.60 (1 − 0.6)5−0
𝑃(0) = 0.01024
𝑃(1) = 5𝐶1 ∗ 0.61 ∗ (1 − 0.6)5−1
𝑃(1) = 0.0768
𝑃(2) = 5𝐶2 ∗ 0.62 ∗ (1 − 0.6)5−2
𝑃(2) = 0.2304
𝑃(3) = 5𝐶3 ∗ 0.63 ∗ (1 − 0.6)5−3
𝑃(3) = 0.3456
= 0.3456 + 0.2304 + 0.0768 + 0.01024
= 𝟎. 𝟔𝟔𝟑𝟎𝟒
III)
Probability that at least one favour the ban
𝑃 = 1 − 𝑃(0)
Take note that P (0) has been calculated already and it is equal to 0.01024
1 − 0.01024
= 𝟎. 𝟗𝟖𝟗𝟕𝟔
QUESTION TWO C
I)
𝑃(𝐴 < 2) = 𝑃(𝐴 ≤ 1)
𝑃(𝐴 ≤ 1) = 𝑃(𝐴 = 0) + 𝑃(𝐴 = 1)
We use the following formula to find the probabilities
𝜆𝑥 𝑒 −𝜆
𝑥!
Where 𝜆 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 = 2.5 and let x=number of accidents
𝒙=𝟎
2.50 𝑒 −2.5
𝑃(𝑥 = 0) =
0!
𝑃(𝑥 = 0) = 0.08208
𝒙 = 𝟏,
𝑃(𝑥 = 1) =
2.51 𝑒 −2.5
1!
𝑃(𝑥 = 1) = 0.2052
Therefore,
𝑃(𝑥 < 2) = 0.08208 + 0.2052
= 0.2873
II)
𝑃(𝑥 > 2) = 1 − 𝑃(𝑥 ≤ 2)
1 − [𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2)
Since the probability of 0 and 1 have already been calculated in (1), the probability of 2 is as
follows
𝑃(𝑥 = 2) =
2.52 ∗ 𝑒 −2.5
2!
𝑃(𝑥 = 2) = 0.2565
Therefore,
𝑃(𝐴 > 2) = 1 − [0.08208 + 0.2052 + 0.2565)
1 − 0.54378
𝑷(𝒙 > 𝟐) = 𝟎. 𝟒𝟓𝟔𝟐𝟐
III)
4
2.50 𝑒 −2.5
𝑃(0 𝑖𝑛 𝑓𝑜𝑢𝑟 𝑤𝑒𝑒𝑘) = [
]
0!
= 𝟎. 𝟎𝟎𝟎𝟎𝟒𝟓𝟑𝟗𝟗
QUESTION 3 A
I) Mean weight Per Kg
𝑀𝑒𝑎𝑛 = 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑀𝑒𝑎𝑛
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑚𝑒𝑎𝑛 = ∑ 𝑤𝑒𝑖𝑔ℎ𝑡 𝑖𝑛 𝑏𝑎𝑔𝑠 ∗ 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑎𝑔𝑠
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑚𝑒𝑎𝑛
= (44 ∗ 0.04) + (45 ∗ 0.15) + (46 ∗ 0.21) + (47 ∗ 0.29) + (48 ∗ 0.20)
+ (49 ∗ 0.10) + (50 ∗ 0.05)
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑚𝑒𝑎𝑛 = 46.9𝑘𝑔𝑠
II)
Standard deviation
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝐸𝑃(𝑤𝑒𝑖𝑔ℎ𝑡 − 𝑚𝑒𝑎𝑛 𝑤𝑒𝑖𝑔ℎ𝑡)2
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 0.04(44 − 46.9)2 + 0.13[45 − 46.9]2 + 0.21[46 − 46.9]2 + 0.29[47 − 46.9]2
+ 0.20[48 − 46.9]2 + 0.10[49 − 46.9]2 + 0.03[50 − 46.9]2
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 2.0553
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
= √2.0553
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑎𝑖𝑡𝑖𝑜𝑛 = 1.4336𝑘𝑔𝑠
III)
Mean cost of producing mealie meal
𝐶 = 75 + 2𝑋
We are going to replace the value of x with 46.5 in the cost function above to find the mean cost.
𝑥 = 46.9
𝐶 = 75 + 2(46.9)
𝐶 = 168.8
Standard deviation cost of producing a bag of mealie-meal
Standard deviation = 1.4336
Where there is x in the cost function it will be replace with the standard deviation
𝐶 = 75 + 2(1.4336)
𝐶 = 77.8672
QUESTION THREE B
𝑖)𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)
= 0.17 + 0.46
= 𝟎. 𝟔𝟓
𝑖𝑖)𝑃(𝐴′ ) = 1 − 𝑃(𝐴)
= 1 − 0.17
= 𝟎. 𝟖𝟑
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑖𝑖𝑖)𝑃(𝐴
∪ 𝐵)
= 1 − 𝑃(𝐴) ∗ 𝑃(𝐵)
1 − 0.0782
= 0.0.9218
IV)
𝑃(𝐴̅𝑈𝐵̅ ) = 1 − 𝑃(𝐴)
= 1 − 0.17 ∗ 0.46
1 − 0.0782
= 0.9218
V)
𝑃(𝐴̅𝑛𝐵̅ ) = 1 − 𝑃(𝐴) + 𝑃(𝐵)
= 1 − 0.63
= 0.37