Answer to Question #114651 in Statistics and Probability for Hanan Hamid

Question

Question 5:
A box contains 4 bad and 6 good bulbs. Two are selected randomly. One of them is tested and found good. What is the probability that other will be good too?
“A” student in a class can solve 75% of the problems and another student “B” can solve 70%. What is the probability either A or B can solve a problem if chosen at random?

Question 6:
Let A and B are two events such that P(A)= ¼, P(A/B)=1/2, P(B/A)=2/3
Let A and B are independent events
A & B are mutually exclusive events.
Find P(A∩B) and P(B)

Question 7:
Two coins are tossed. What is the conditional probability that two heads result, given that there is at least one head?

Question 8:
Two fair dice, one red and one green, are thrown. Let A denote the event that the red die shows an even number and B, the event the green die shows 5 or 6. Show that event A & B are independent.
[hint: you may take help from last file of probability handouts there mentioned formulas of dependent/independent events]

Answer 1

Question 5: 

Sample space for two bulbs

$(good, good), (good, bad), (bad, good), (bad, bad)$

$$P(good, good)={\dbinom{6}{2}\dbinom{4}{0} \over \dbinom{10}{2}}={1 \over 3}$$

$$P(good)={6 \over 10}={3 \over 5}$$

$$P(good|good)={P(good, good) \over P(good)}={{1 \over 3} \over {3 \over 5}}={5 \over 9}$$

$$P(A)=0.75, P(B)=0.7$$

$$P(A^C)=1-P(A)=1-0.75=0.25$$

$$P(B^C)=1-P(B)=1-0.7=0.3$$

$$P(A^C\cap B^C)=P(A^C)\cdot P(B^C)=0.25\cdot 0.3=0.075$$

$$P(A\cup B)=1-P(A^C\cap B^C)=1-0.075=0.925$$

Question 6:

Let A and B are two events such that P(A)= ¼, P(A/B)=1/2, P(B/A)=2/3

Let A and B are independent events

A & B are mutually exclusive events.

Find P(A∩B) and P(B) 

$$P(A|B)={P(A\cap B) \over P(B)}, P(B|A)={P(A\cap B) \over P(A)}$$ $$P(A)=1/4, P(A|B)=1/2, P(B|A)=2/3$$

$$P(A\cap B)=P(A)P(B|A)={1 \over 4}\cdot{2 \over 3}={1 \over 6}$$

$$P(B)={P(A\cap B) \over P(A|B)}={ {1 \over 6}\over {1 \over 2}}={1 \over 3}$$

Independent events: $P(A\cap B)=P(A)P(B).$

$$P(A|B)=P(A), P(B|A)=P(B)$$

Mutually exclusive events: $P(A\cap B)=0$

$$P(A|B)=P(B|A)=0$$

Question 7: 

Sample space for two coins

$$(HH), (HT), (TH), (TT)$$

$$P(HH|at\ least \ H)={P(HH) \over P(at\ least \ H)}={ {1 \over 4}\over {3 \over 4}}={1 \over 3}$$

Question 8:

Sample space for two dice

$(R1, G1), (R1, G2), (R1, G3), (R1, G4), (R1, G5), (R1, G6),$

$(R2, G1), (R2, G2), (R2, G3), (R2, G4), (R2, G5), (R2, G6),$

$(R3, G1), (R3, G2), (R3, G3), (R3, G4), (R3, G5), (R3, G6),$

$(R4, G1), (R4, G2), (R4, G3), (R4, G4), (R4, G5), (R4, G6),$

$(R5, G1), (R5, G2), (R5, G3), (R5, G4), (R5, G5), (R5, G6),$

$(R6, G1), (R6, G2), (R6, G3), (R6, G4), (R6, G5), (R6, G6),$

$$P(A)={18 \over 36}={1 \over 2}$$

$$P(B)={12 \over 36}={1 \over 3}$$

$$P(A\cap B)={6 \over 36}={1 \over 6}={1 \over 2}\cdot {1 \over 3}=P(A)P(B)$$

The events $A$ and $B$ are independent.