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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 _______________.
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+1
0.462 ?

7 Answers

+30 votes
Best answer

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
edited by
+2
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.
0
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.
+7
@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
No, the answer key is out already.
0

@akash.dinkar12

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

0
NO...then their MSB would become different.
+3
I think the word "independent and uniform" is game-changing.

If the term was not given then,

$\frac{7*6+6*5}{13*12}$ $=$ $\frac{72}{156}$ $=$ $0.4615$
+12 votes
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
edited by
0
Independent Events tha 0.462 should not be the answer
0
Updated !!! I missed that. Thanks ...
0
I also missed that in exam😒
+10 votes

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
+5 votes

Ans will be 0.462 only!!

​​​

by
+3
What I have done --

(7C2  + 6C2)/(13C2)

Got 0.461.
+2 votes

If we do $$\frac{{^7C_2} + {^6C_2}}{{^{13}C_2}}$$

it won't be correct as by doing this we ensure that in choosing 2 elements, they both will necessarily be distinct.

But to be independent, we should be able to select an item n number of times in n trials. Because choosing the same item before shouldn't matter in the present, as the trials are independent. We need to allow repetitions.

 

So,

$$\frac{7}{13}*\frac{7}{13}+\frac{6}{13}*\frac{6}{13}$$

will be correct.

 

0.503 is the answer.

by
+2 votes

$\mathbf{\underline{Answer:\Rightarrow}}\;\;\mathbf{\underline{\bbox[orange, 5px, border:2px solid red]{\color{darkblack}{.5029}}}}$

$\mathbf{\underline{Explanation:\Rightarrow}}$

$\mathbf{\underline{Importance\; of \;word\; \color{blue}{"independently"} \;in\; the\; question:}}$

The word $\underline{\color{blue}{\mathbf{independent}}}$ here means that after selecting a number from the set of numbers, your count of number, that is, the sample space hasn't decreased.

In other words, it can be compared with the problem of picking a ball from the bag and then keeping it again in the bag. Then you can pick the next ball again from the same number of balls.

 

Total numbers with $\mathbf{0}$ as the significant bits $=\mathbf 7$

Total numbers with $\mathbf{1}$ as the significant bits $=\mathbf 6$

Now,

The probability of picking the number with same $\mathbf{MSB-0} =\mathbf{ \dfrac{7C_1\times7C_1}{13\times13}}$

The probability of picking the number with same $\mathbf{MSB-1 = \dfrac{6C_1\times6C_1}{13\times13}}$

$\therefore$ Total probability $\mathrm{=\dfrac{7C_1\times7C_1}{13\times13}+\dfrac{6C_1\times6C_1}{13\times13} = \dfrac{49}{169}\times \dfrac{36}{169} = \mathbf{\underline{\bbox[orange, 5px, border:2px solid red]{\color{darkblack}{.5029}}}}}$

$\mathbf{\color{blue}{\underline{Binary\;representation\;of\;Numbers:}}}$

$\mathbf{NUMBER}$ $\mathbf{MSB}$     $\mathbf{LSB}$
0 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 0 0 0
1 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 0 0

1

2 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 0 1 0
3 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 0 1 1
4 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 1 0 0
5 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 1 0 1
6 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 1 1 0
7 $\bbox[yellow, 5px, border:2px solid red]{\color{darkorange}0}$ 1 1 1
8 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 0 0 0
9 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 0 0

1

10 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 0 1 0
11 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 0 1 1
12 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 1 0 0
13 $\bbox[blue, 5px, border:2px solid red]{\color{darkorange}1}$ 1 0 1

 

$\therefore$ The correct answer is $\mathbf{.5029}$

by
+1 vote

Answer : 0.503

the word "chosen independently and uniformly at random" makes big impact on approach,

0001 , 0010 , 0011 , 0100 , 0101 , 0110 , 0111   total=7  :- have 0 at MSB

1000 , 1001 , 1010 , 1011 , 1100 , 1101              total=6 :- have 1 at MSB

(C(7,1)*C(7,1) + C(6,1)*C(6,1)) / ( C(13,1) * C(13,1) )

= ((7*7)+(6*6)) / (13*13) = 0.503

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