Midterm

1. True

2. True

3. False

4. False

5. False

6. True : probability distribution definition:

Is an assignment of a number P(si), the probability of si, to each outcome of a finite sample space S={s1,s2, …. , sn}

* Must satisfy 0~1

* Everything combines to be 1

7. If you roll a pair of undistinguishable for dice, then the probability of the event A={(1,2), (2,2), (6,3)}= 3/21

8. False

v, e, e, k, e

5*4*1=20

(5*4*3*2*1)/(3*2*1) =20

Problem 2.

20 contestants

a. 20! = 20*19*18*17*16*15*14*13*12*11*…

b. Picking 3 out of 20. C(20,3) = (20*19*18)/(3*2*1)=6830/6=1140

Problem 3.

36% dog, 30% cat, 22% (Cat|Dog) = of the ppl who have a dog, 22% have a cat

a. Both? = 0.22*0.30 = 0.066

b. (Dog|Cat)=? [P(Cat|Dog)*P(Dog)] / [P(Cat|Dog)*P(Dog)+P(Cat|Dog’)*P(Dog’)

= (0.22*0.36)/(0.22*0.36+0.78*0.64) = 0.0792/ (0.0792+0.4992) = 0.0792/0.5784= 0.1591

P(A|T)= [P(T|A)*P(A)]/[P(T|A)P(A)+P(T|A’)*P(A’)

Problem 4.

150 employees. 80=women; 77=unionized; 86=married; 47=unionized women; 52= married women; 45= unionized & married; 34 unionized married women

Problem 5.

12 Herb; 8 Fungi

Took 6 random

1. Probability that 6=herb?

(12/20)*(11/19)*(10*18)*(9*17)*(8*16)*(7*15)

665280/ 27907200

= 0.02383 = 2.38%

* Also think of it as

C(12,6)/C(20,6)

= (choosing 6 herb bottles out of herb 12)/(choosing 6 bottles out of 20)

2. Probability that at least 2=herb?

Probability that 0 or 1 herb

1-{(all fungi: 8/20)*(7/19)*(6/18)*(5/17)*(4*16)*(3/15) - 20160/27907200

+ (1 herb and 5 fungi: 12/20)*(8/19)*(7/18)*(6/17)*(5/16)*(4/15)} -80640/27907200

-> 1- (100800/27907200) = 1- 0.003611971104… = 0.9963?

To choose 6 fungi out of 8

To choose 5 fungi out of 8

Divided by all

Problem 6. 5 red balls, 3 blue balls, 2 yellow balls

a.

i) the ball is red: 5/10=50%

ii) the ball is either blue of yellow: 5/10=50%

b. one-by-one, without replacement

i) both balls...