For a departmental inquiry committee, an employer randomly selects a sample of three employees from ten employees, six men and four women. If the selection is at random, what then is the probability of selecting at least one-man member employee?
Solution
A married couple wants to enjoy the weather of hill stations at Murree and Abbottabad. The probabilities of enjoying weather at Murree and Abbottabad are respectively 0.50 and .12.
What is the probability that a couple will enjoy at least one of these two hill stations?
TREE DIAGRAM
Event-1
= Weather of hill station at Murree
Event-2
= Weather of hill station at Abbottabad
P(E1)
= 0.5
P(E2)
= 0.12
P’(E1)
= 1 – 0.5
= 0.5
P’(E2)
= 1 – 0.12
= 0.88
P(E1
Ռ
E2)’ = 0.5 * 0.88
= 0.44
P(
couple will enjoy at least one of these two hill stations) = 1- 0.44
= 0.56
= 56%
What concept of the events do
you give to these two hill stations?
We use the concept the mutually
exclusive events. Because these two events can’t occurred at a time.
Q# What do you mean by objective and subjective probabilities? Fit one simple real-life example in each case, which really reflect its linkage to objective and subjective probabilities.
Objective
probabilities
Objective
probabilities are based on mathematical analysis, experiments, and mathematical
equations than anecdotes, personal experience, or hunting. In the financial
world, taking advantage of opportunities can be very important to prevent from
making emotional decisions when investing.
We often deceive ourselves into thinking
that "we have always been lucky in car investments" or that "we
have not lost money on gold".
Subjective
Probabilities
The probabilities
where we use our ideas, emotions and thoughts on the basis of past experience
to find opportunities and probabilities. Example: we think,we have an 80%
chance that our best friend will call today, because her car broke down
yesterday and he will need a ride.
Example
let's say Ali buys a raffle ticket to
support the local Girl Scouts home team. The team sells 1,000 tickets. From a
standpoint, Ali has 1 in 1,000 chance of winning. But humbly, John thinks his
chances of winning are very high because he "feels good about it."
However, his chances are still 1 in 1,000.
b. In a survey of 200 college students, it was found that;
120 study mathematics, 90 study physics
70 study chemistry, 40 study mathematics
and physics,
30 study physics and
chemistry, 50 study chemistry and
mathematics
20 study none of these
subjects.
Depict the events “A: Mathematics”, “B: Physics” and “C: Chemistry” on the given Venn-Diagram, and find the number of the students who studies all the three subjects.
Solution#1
Let Assume
A =
Mathematics ; B = Physics and C = Chemistry
n(A) = 120 , n(B)
= 90 ,
n(C) = 70
n
( A ∩ B) = 40 ,
n ( B ∩ C ) = 30
n
( C ∩ A ) = 50 , n ( A∪B∪
C )’ = 20
Now
n(A∪B∪
C)’ = n(U) – n(A∪B∪ C)
20
= 200 – n (A∪B∪ C)
Therefore,
n(A∪B∪ C) = 200 – 20 = 180
n(A∪B∪
C)
=
n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
180
= 120 + 90 + 70 - 40 - 30 - 50 + n(A ∩ B ∩ C)
⇒ n(A ∩ B ∩ C) =180 - 120 - 90 - 70 + 40 + 30 +
50
⇒ n(A ∩ B ∩ C)
= 20. (Students who studies all three subjects)
Q# What did you learn from binomial and Hyper- geometric distributions? Write a brief note of five lines on these distributions.
Answer
A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times. The binomial is a type of distribution that has two possible outcomes (the prefix “bi” means two, or twice). For example, a coin toss has only two possible outcomes: heads or tails and taking a test could have two possible outcomes: pass or fail.
For example, let’s suppose we
wanted to know the probability of getting a 1 on a die roll. if you were to
roll a die 20 times, the probability of rolling a one on any throw is 1/6. Roll
twenty times and you have a binomial distribution of (n=20, p=1/6). SUCCESS
would be “roll a one” and FAILURE would be “roll anything else.” If the outcome
in question was the probability of the die landing on an even number, the
binomial distribution would then become (n=20, p=1/2). That’s because your
probability of throwing an even number is one half.
The hypergeometric distribution is a discrete distribution that models the number of events in a fixed sample size when you know the total number of items in the population that the sample is from. Each item in the sample has two possible outcomes (either an event or a nonevent). The samples are without replacement, so every item in the sample is different. When an item is chosen from the population, it cannot be chosen again. Therefore, an item's chance of being selected increases on each trial, assuming that it has not yet been selected.
Use
the hypergeometric
distribution for samples that are drawn from relatively small populations,
without replacement. For example, the hypergeometric distribution is used
in Fisher's exact test to test the difference between two proportions, and in
acceptance sampling by attributes for sampling from an isolated lot of finite
size.
The
hypergeometric distribution is defined by 3 parameters: population size, event
count in population, and sample size.
For
example, We receive one special order shipment of 500 labels. Suppose that 2%
of the labels are defective. The event count in the population is 10 (0.02 *
500). we sample 40 labels and want to determine the probability of 3 or more
defective labels in that sample. The probability of 3 of more defective labels
in the sample is 0.0384.
Q# What did you learn from binomial and Hyper- geometric distributions? Write a brief note of five lines on these distributions.
Answer
A binomial
distribution can be thought of as simply the
probability of a SUCCESS or FAILURE outcome in an experiment or survey that is
repeated multiple times. The binomial is a type of distribution that has two possible outcomes (the
prefix “bi” means two, or twice).
For example, a coin toss has only two possible outcomes: heads or tails and
taking a test could have two possible outcomes: pass or fail.
For example, let’s suppose we
wanted to know the probability of getting a 1 on a die roll. if you were to
roll a die 20 times, the probability of rolling a one on any throw is 1/6. Roll
twenty times and you have a binomial distribution of (n=20, p=1/6). SUCCESS
would be “roll a one” and FAILURE would be “roll anything else.” If the outcome
in question was the probability of the die landing on an even number, the
binomial distribution would then become (n=20, p=1/2). That’s because your
probability of throwing an even number is one half.
The hypergeometric distribution is a discrete distribution that models the number of events in a fixed sample size when you know the total number of items in the population that the sample is from. Each item in the sample has two possible outcomes (either an event or a nonevent). The samples are without replacement, so every item in the sample is different. When an item is chosen from the population, it cannot be chosen again. Therefore, an item's chance of being selected increases on each trial, assuming that it has not yet been selected.
Use
the hypergeometric
distribution for samples that are drawn from relatively small populations,
without replacement. For example, the hypergeometric distribution is used
in Fisher's exact test to test the difference between two proportions, and in
acceptance sampling by attributes for sampling from an isolated lot of finite
size.
The
hypergeometric distribution is defined by 3 parameters: population size, event
count in population, and sample size.
For
example, We receive one special order shipment of 500 labels. Suppose that 2%
of the labels are defective. The event count in the population is 10 (0.02 *
500). we sample 40 labels and want to determine the probability of 3 or more
defective labels in that sample. The probability of 3 of more defective labels in
the sample is 0.0384.
b.
Machines A, B and C make resistors 1, 2 and 3 respectively. The
acceptability of the three resistors are 90%, 93% and 81% respectively. Three
resistors arepicked at random sampling with replacement. Find the probability
that
i.
All
resistors are acceptable.
ii.
P(A) = 0.9
iii.
P(B) = 0.93
iv.
P(C) = 0.81
Since
A , B and C are the independent events, therefore
P(AՌBՌC)
= P(A) P(B) P(C)
= 0.9 * 0.93 * 0.81
= 0.678
= 67.8% Answer
v.
All
resistors are not acceptable.
P(all
resistors are not acceptable) = 1
- P(all resistors are acceptable)
= 1 – 0.678
= 0.322
= 32.2% Answer
vi.
At
least one resistor is acceptable. [Marks-01]
P(At least one resistor is acceptable) =
P(AUBUC)
P(AUBUC)
= P(A) +P(B) +P(C) -P(AՌB)-P(BՌC) –
P(AՌC) + P(AՌBՌC)
= 0.9 + 0.93 +0.81 – 0.9*0.93 – 0.93*0.81 –
0.9*0.81 + 0.678
= 0.998
= 99.8% Answer
Q# Bay’s Theorem: A Construction Company finds that an experienced machine operator of its company (one or more years of experience) will produce a defective item 1.5% of the time. Operators with some experience (up to one year) will produce a 2.5% defective rate and new operators will produce a 5% defective rate. The company has 65% experienced employees, 25% with some experience and 10% new employees. Find the probability that
i.
Experienced operator of a company
produced a particular defective item.
ii. Operator with some experience of a company produced a particular defective item.
iii.
New operator of a company produced a
particular defective item.
let’s create some events to work with.
D: “operator produces
a defective item”
E: “experienced
operator”
S: “operator with some experience”
N: “new operator”
With these definitions, the information in the problem
statement can be written as
P(E) = 0.65
P(D/E) = 0.015
P(S) = 0.25
P(D/S) = 0.025
P(N) = 0.1
P(D/N) = 0.05
The
probability that an experience operator produces a defective item, we are
interested in the event E and D. These probabilities lie
along the top branch so
P(E
and D) = P(D/E) P(E) = (0.015)(0.65)
= 0.0097
=0.97% Answer
To find
the probability that a particular defective item was produced by a Experienced
operator, we need to compute P(E | D). The appropriate form
of Baye’s Theorem is
=
= 0.4642
= 46.42% Answer
To find
the probability that a particular defective item was produced by a Some Experienced
operator, we need to compute P(S | D). The appropriate form
of Baye’s Theorem is
=
= 0.2976
= 29.76% Answer
To find the probability that a particular
defective item was produced by aNew operator, we need to compute P(N | D). The appropriate form
of Baye’s Theorem is
P
=
= 0.2381
= 23.81% Answer
