STA301 Current Final Term Fall 2013 Shared by ihsan

P(A) , P(B) and P(AnB) was given we have to find P(AUB).
P(A)=13/52
P(B)=12/52
P(A ∩ B) =3/52
P(AUB)=?
Sol: By addition theorem of probability we have
P(AUB)=P(A)+P(B)- P(A ∩ B)
So P(AUB)=13/52+12/52-3/52 = 13+12-3/53=22/52
 Note: Same example is given on Page # 152 of handouts
EXAMPLE
If one card is selected at random from a deck of 52 playing cards, what is the probability that the card is a club or a face
card or both?
Let A represent the event that the card selected is a club, B, the event that the card selected is a face card, and A ∩ B,
the event that the card selected is both a club and a face card. Then we need P (A ∪ B)
Now P(A) = 13/52, as there are 13 clubs, P(B) = 12/52, as there are 12 faces cards,
P(A ∩ B) = 3/52, since 3 of clubs are also face cards.
Therefore the desired probability is
P (A ∪B) = P (A) + P (B) – P (A ∩ B)
= 13/52 + 12/52 - 3/52
= 22/52.
Another q from current papers
Given that
P(A)=13/52, P(B)=12/52, P(A∩B)=3/52 ,    FIND P(A|B)?

Sol:
As by conditional probability; we know that
P(A/B)= P(A∩B)/P(B)
So;
P(A/B)= (3/52)/(12/52)
P(A/B)= (3/52)x(52/12)
P(A/B)= (3/12)
P(A/B)= (1/4)

Another q & its solution from current exam
For a certain data: (3 marks)
Mean = 3.8
Median = 2.3
and Standard Deviation = 1.3
Find Pearson’s coefficient of skewness

Solution
As we know that
Find Pearson’s coefficient of skewness
= (mean-mode)/standard deviation = 3(mean-median)/sd.
=3(3.8-2.3)/1.3
= 3(1.5)/1.3
= 4.5/1.3
=3.4615