Question

A bag contains 18 blocks numbered from 1 to 18. two blocks are selected from the bag at the same time. let event e = the two numbers differ by 11. how many outcomes are in e?
a.7
b.14
c.28
d.292

Answer 1

probable cases are   1, 12  or 2,13 or 3,14 or 4,15, or 5, 16 or 6, 17 or 7,18

and reverse

so, total favourable cases = 14

Answer 2

Answer:  The correct option is (A) 7.

Step-by-step explanation:  Given that a bag contains 18 blocks numbered from 1 to 18. Two blocks are selected from the bag at the same time.

Let, E = even that the two numbers differ by 11.

Is 'S' is the complete sample space for the event of selecting two blocks, then

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), . . , (17, 18), (18, 18)}.

The number of elements of the sample space 'S' is

$n(S)=^{18}C_2=\dfrac{18!}{2!(18-2)!}=\dfrac{18\times 17\times 16!}{2\times 1\times 16!}=153.$

The event 'E' is a subset of the above sample space 'S'.

We are to find the number of outcomes of the event 'E'.

Since 'E' is the event such that the two numbers differ by 11, so

E = {(1, 12), (2, 13), (3, 14), (4, 15), (5, 16), (6, 17), (7, 18)}.

Therefore, n(E) = 7. That is, there are 7 outcomes in event 'E'.

Thus, the correct option is (A). 7.