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16 Jun 2018, 00:16
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:12) correct
25%
(02:17)
wrong
based on 267
sessions
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9 balls labelled 1 - 9 are placed within a jar and two are pulled at random without replacement. What is the probability that the sum of the numbers on those two balls will be even?
A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9
A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9
16 Jun 2018, 01:58
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Bunuel
9 balls labelled 1 - 9 are placed within a jar and two are pulled at random without replacement. What is the probability that the sum of the numbers on those two balls will be even?
A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9
A. 16/81
B. 25/81
C. 4/9
D. 41/81
E. 5/9
We can get Even sum in two cases
ODD + ODD = EVEN
EVEN + EVEN = EVEN
Case 1) Two ODD numbers are picked
P(First Number to be ODD) and P(2nd Number be ODD = \(\frac{5}{9}*\frac{4}{8}\) = \(\frac{20}{72}\)
Case 2) Two Even number are selected
P(First Number to be EVEN) and P(2nd Number be EVEN = \(\frac{4}{9}*\frac{3}{8}\) = \(\frac{12}{72}\)
Total Probability : \(\frac{20}{72}+\frac{12}{72}\) = \(\frac{4}{9}\)
General Discussion
16 Jun 2018, 01:52
Kudos
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For the sum to be even both the numbers should be odd or even.
Odd numbers (1,3,5,7,9) = 5 numbers
Even Numbers (2,4,6,8) = 4 numbers
Odd + odd will be: 5 * 4
Even + even will be: 4 * 3
Selecting two numbers will be: 9 * 8
So, probability of selecting two numbers = ((5*4) + (4*3))/(9*8) = 4/9
OA - C.
Odd numbers (1,3,5,7,9) = 5 numbers
Even Numbers (2,4,6,8) = 4 numbers
Odd + odd will be: 5 * 4
Even + even will be: 4 * 3
Selecting two numbers will be: 9 * 8
So, probability of selecting two numbers = ((5*4) + (4*3))/(9*8) = 4/9
OA - C.
