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

edited ago | 1k views
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+6
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
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What is the meaning of uniformly at random?

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

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

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

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.

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

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 Boss (18.8k points)
selected ago
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@Ram Swaroop

what was the problem in the answer.

Why do you deselect??

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why is denominator multiplied with 13*13?
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Because of the word "independent"
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 (597 points)
+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)
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all are saying o.508 because of independence word
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I don't know how because I am getting 0.4615
by (41 points)