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Two numbers are chosen independently and uniformly at random from the set $\{1,2,\ldots,13\}.$

The probability (rounded off to 3 decimal places) that their $4-bit$ (unsigned) binary representations have the same most significant bit is _______________.

edited | 4.8k views
+1
0.462 ?

These are two groups:

1. MSB with 1
2. MSB with 0

$n_{MSB0} = 7, n_{MSB1} = 6, n_{total}=13$

Choose randomly and INDEPENDENTLY two elements out of 13 elements such that MSB is same.

$P = \frac{n_{MSB1}*n_{MSB1} + n_{MSB0}*n_{MSB0}}{n_{Total}}$

$P = \frac{7*7+ 6*6}{13*13} = \frac{85}{169} = 0.5029$

by Veteran (60.5k points)
edited by
+1
Yes sir correct!! upto 3 decimal places ans will be 0.503
0
I got 0.515

Will they give marks???
0
Sir but in this case order will matter.

i.e. If we select 2 and 1 then 2,1 and 1,2 will be treated as different selections.
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Can someone tell me, why taking MSB1 from 7 bits
+1

@srestha

write down numbers in binary u will get it..

+5
You don't select both at a time. You select one after the other. So msb1 should be from 7 and then 6 while for msb0 it should be from 6 and then 5. Isn't it so? This way the answer is 0.462.
+4
@kajal sanklecha answer is correct because it is saying independent (not dependent) mean if we choose 1 no. we can choose same no.again and uniform distribution is sufficient to say that probability of sample space is 1 i.e if 1<=n<=13 Probability is 1 otherwise 0.
+1
Yes due to the word independent ans is 0.503
0
0

@akash.dinkar12

is it not possible one number is taken from MSB=0 and another number taken from MSB=1?

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NO...then their MSB would become different.
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I think the word "independent and uniform" is game-changing.

If the term was not given then,

$\frac{7*6+6*5}{13*12}$ $=$ $\frac{72}{156}$ $=$ $0.4615$
No of choices with MSB 0 = ( 1, 2, 3, 4, 5, 6, 7}

No of choices with MSB 1 = ( 8, 9, 10, 11, 12, 13}

Since independently assumption is made , Probability = (7*7 + 6*6) / 13*13 = 0.503

The choosing of a number from the set doesn't depend upon the previous selection.
by Active (2.6k points)
edited
0
Independent Events tha 0.462 should not be the answer
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Updated !!! I missed that. Thanks ...
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I also missed that in exam😒

Ans will be 0.462 only!!

​​​

by Junior (893 points)
+3
What I have done --

(7C2  + 6C2)/(13C2)

Got 0.461.

Two events are said to be independent of each other if the probability of occurrence of the first event does not affect the probability of occurrence of the second event.

So, here it means that whenever we choose one element from the set of 7 elements with MSB 0 for the first time, we can choose one more element from those 7 elements for the second time which gives us 7*7=7^2 favourable outcomes. Similarly, for the case of set of 6 elements with MSB 1, we get 6^6=6^2  more favourable outcomes. So,total favourable outcomes = 7^7 + 6^6 = 85 and the final answer is 85/13^2=0.502.

PS: I too kept the answer as (7C2  + 6C2)/(13C2) = 0.461 in Gate exam but realized the mistake later.

by (71 points)