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Given Set $A= {2, 3, 4, 5}$ and Set $B= { 11, 12, 13, 14, 15}$, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals $16$?

1. $0.20$
2. $0.25$
3. $0.30$
4. $0.33$

edited | 1.5k views

option A because total combinations are $5\times 4=20$ and out of $20$ we have only $4$ combinations which have sum $16$

1. $2,14$
2. $3.13$
3. $4.12$
4. $5,11$
by Active (5k points)
edited by
0
It is possible cases or not?

(2,14)  (14,2)
(3.13)  (13,3)
(4.12)  (12,4)
(5,11)  (11,5)
0
buddy 2+14 = 14+2

so we consider it , one time only.
+2

(2,14)  (14,2)
(3.13)  (13,3)
(4.12)  (12,4)
(5,11)  (11,5)

This will also be correct

8 / 40 ==>0.2

0
yes ,set does not contain duplicate
+1 vote

option A

probability =Number of favorable case/total case

The favorable cases are:-

2,14
3.13
4.12
5,11

4C1=4

total case =A number selected from set A * A number selected from set B

4C1*5C1=20

probability =Number of favorable case/total case=4/20=0.20

by Active (3.3k points)
+1 vote
After choosing an element from A, there is only one element in B out of 5 elements that makes sum 16.

So, 1/5 = 0.2
by (273 points)

1