PROBABILITY

Probability

Probability

1) A group contains 4 men and 6 women, among them Max Brown and Sue Green. If  3 are randomly selected from the group, find: (a) P(all 3 are women); (b) P(there is at

 least 1 man); (c) P(Both Max and Sue are chosen).

n= 4M + 6W= 10

a) P(all 3 are women)

P(all 3 are women)= = 0.1666

(b) P(there is at least 1 man)

P(there is at least 1 man)= =

=P(there is at least 1 man)==0.3

(c) P(Both Max and Sue are chosen)

P(Both Max and Sue are chosen)=

=0.01

2. a) If P(A) = 5/7, what are the odds against A? In favor of A?

The odds of an event occurring is the ratio of the number of ways the event can occur (successes) to the number of ways the event cannot occur (failures).

Odd against A= 2:5

In favour of A= 5:2

b)

We have to calculate p(n) where p(n) is the probability of at least two of the 4 people sharing a birthday on the same day of the week

P(n)= 7 !/(7^4(7-4)!=0.34

3. Consider the 5-coin toss experiment. Construct a probability table for the random var- iable X, the number of heads. Find f(x), F(x), the mean E(x), and the variance E((x-E(x)^2), and interpret each.

Possible outcomes= =32

(HHHHH)

(HHHHT)

(HHHTH)

(HHTHH)

(HTHHH)

(THHHH)

(HHHTT)

(HHTTH)

(HTTHH)

(TTHHH)

(HHTHT)

(HTHHT)

(THHHT)

(HTHHT)

(THHTH)

(THTHH)

(HHTTT)

(HTTTH)

(TTTHH)

(HTHTT)

(HTTHT)

(THTHT)

(TTHHT)

(TTHTH)

(HTTTT)

(THTTT)

(TTHTT)

(TTTHT)

(TTTTH)

(THHTT)

(HTHTH)

(TTTTT)

f(x) =p(x) x= 0,1,2,3,4,5

P(one head)= 5/32

P(occurring two head)=10/32

P(occurring three head)=10/32

P(occurring four head)=5/32

P(occurring five head)=1/32

P(occurring no head)=1/32

F(x)=

For E (x)

E(x)=

E(x)=0*(1/32)+ 1*(10/32) + 2*(10/32) +3 (10/32) + 4*(5/32) + 5*(1/32)

= 2.65

For V(x)

V(x)= E() - (E(x))^2



E()= =0*(1/32)+ 1*(10/32) + 4*(10/32) +9* (10/32) + 16*(5/32) + 25*(1/32)

=7.65

V(x)= E() - (E(x))^2

Hence

V(x)= 7.65-2.65

V(x)=5

4. A box contans 4 red, 3 yellow and 5 blue cubes, all of the same size and shape. If 3 are randomly selected from the box, find: (a) P(2 red and 1 yellow); (b) P(all 3 blue).

  Answer EACH question TWICE, once WITH replacement and once WITHOUT.

P( 2 Red and 1 Yellow)

Without Replacement

Solution: Let A1 = the event that the first cube is black; A2 = the event that the second marble is Red and A3: = the event that the third cube Yellow

In the beginning, there are 12 cubes in the box, 4 of which are Red. Therefore P(A1)=4/12

After the first selection, there are 11 cubes in the box, 3 of which are red. Therefore, P(A2|A1)= 3/11

After the second selection, there are 10 cubes in the box, 3 of which are Yellow therefore, P(A3| A1nA2)=3/10

Therefore, based on the rule of multiplication:

P (A1nA2nA3) = P(A1) P(A2|A1) P(A3| A1nA2)= 4/12 *3/11* 3/10

P (A1nA2nA3)=0.027

With Replacement

We repeat the experiment; but this time we select cubes with replacement. That is, we select one cube, note its color, and then replace it in the box before making the second selection and the third selection:

Solution: Let A1 = the event that the first cube is black; A2 = the event that the second marble is Red and A3: = the event that the third cube Yellow

In the beginning, there are 12 cubes in the box, 4 of which are Red. Therefore P(A1)=4/12

After the first selection, we replace the selected cube; so there are still 12 ...