Chapter 11I11-1. (a) 10/15 (b) 3/15 (c) 2/15.
I11-2. (a) (10/15)(3/14) (b) (10/15)(9/14) (c) (2/15)(3/14)
I11-3.
Method I Label the fruit A1,A2,A3,O1,O2,O3,O4,O5,P1,P2
Next collect all possible pairs — Note the pair (A1,A2) is thesame as the pair
(A2,A1) These are all possible event pairs(A1,A2) (A1,A3) (A1,O1) (A1,O2) (A1,O3) (A1,O4) (A1,O5) (A1,P1) (A1,P2) 9
(A2,A3) (A2,O1) (A2,O2) (A2,O3) (A2,04) (A2,05) (A2,P1) (A2,P2) 8
(A3,01) (A3,02) (A3,03) (A3,04) (A3,05) (A3,P1) (A3,P2) 7
(O1,02)(O2,03) (O1,03)(02,04) (O1,04)(O2,05) (O2,P1)(O1,05) (O1,P1)(O2,P2) (O1,P2) 56
(03,04) (03,05) (O3,P1) (O3,P2) 4
(O5,P1)(04,05) (O4,P1)(O5,P2) (O4,P2)^32
(P1,P2) 1
Total of 45 possible two event pairs. Assign an equal probability to each event
of 1/45
Since the event (P1,P2) can only happen once —its probability is 1/45
The probability of getting two oranges is 10/45 since there are 10 (0,0) pairs above
The probability of getting two apples is 3/45 since there are3 (A,A) pairs above.
Method 2 From AAA, OOOOO, PP assign a probability of 1/10 to each single
event of selecting one fruit. (All have same probability)
Probability of getting a pear is 1 /10 + 1/10 = 2/10 = 1/ 5
If a pear is selected, then we are left with AAA,OOOOO,P Now each single
event has probability of 1/9
Probability of getting a pear on second selection is 1/9.
The product(1/5)(1/9) = 1/ 45 is probability of getting a pear plus another pear.
Similarly, the probability of getting an orange is 5 /10 = 1/ 2 the first time and
4/9 the second time, so that(1/2)(4/9) = 2/ 9 is probability of getting two oranges.
The probability of getting two apples is(3/10)(2/9) = 6/90 = 1/ 15I11-4. (a) mean 80, variance 79.6, standard deviation 8.922
Solutions Chapter 11