+1 vote
773 views
Q.23
Two numbers are chosen independently and uniformly at random from the set (1, 2,..., 13]. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is

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0
+3
Here the word independently refers after selecting 1st element it won't affect in selecting 2nd element .

The two numbers must have same significant bit . {1,2,3,4,5,6,7} have 0 as MSB and {8,9,10,11,12,13} have 1 as MSB .

But the two numbers may be exactly same or different(having same MSB) .(independent)

So either we can select 1st element as 7C1 and also 2nd element from those 7 element which is 7C1,  because two elements are independent .

Same will happen in case of 6 element having 1 as msb . So the answer will be

((7C1×7C1)+(6C1×6C1))/(13C1×13C1)

=> ((7×7)+(6×6))/(13×13)

=> (49+36)/(169)

=> 85/169

=> 0.502958

As rounding off up to 3 digit the answer will be 0.503

+1 vote
1 to 7 have the most significant digit as 0. Rest 6 have 1 as MSB.

So, $((7*7) + (6 * 6) )/ (13 * 13)$

which is 0.5029
by Junior (551 points)
selected
+1 vote
As there are only 6 numbers in set A out of 13 numbers whose MSB will be 1 ( from 8-13) so the probility is 6/13 i.e. 0.4615
by (119 points)
0
all are saying o.508 because of independence word
0

I don't know how because I am getting 0.4615
by (31 points)