natural numbers from 1 to 10 are written on paper slips and are put in a box. another box contains paper slips with numbers less than 10 which are multiples of 3. one slip is taken from the each box. a- what is the probabily of both being odd ? b- what is the probability of getting at least one even ?

Asked by army29676 | 25th Mar, 2022, 08:35: PM

Expert Answer:

Numbers in first box = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Numbers in the second box = {3, 6, 9}
Total outcomes = {(1, 3), (1, 6), (1, 9), (2, 3), (2, 6), (2, 9), ...., (10, 3), (10, 6), (10, 9)}
Total no. of outcomes = 30
(i)
To find the probabily of both being odd
Favorable outcomes = {(1, 3), (1, 9), (3, 3), (3, 9), (5, 3), (5, 9), (7, 3), (7, 9), (9, 3), (9, 9)}
No. of favorable outcomes = 10
Probability of both being odd = 10/30 = 1/3
 
(ii)
To find the probabily of getting at least one even
Favorable outcomes = {(1, 3), (1, 9), (3, 3), (3, 9), (5, 3), (5, 9), (7, 3), (7, 9), (9, 3), (9, 9)}
No. of favorable outcomes = 10
Probability of getting at least one even = 1 - Probability of both being odd = 1 - 1/3 = 2/3

Answered by Renu Varma | 28th Mar, 2022, 12:38: PM