GATE CSE 2019 | Question: 22

Question

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 $4\text{-bit}$ (unsigned) binary representations have the same most significant bit is ___________.

Comments

Hint guys before you see the solution : INDEPENDENTLY 

Hey if you solved 2019 set 1 and there is one more question which turned evil because of this word INDEPENDENTLY so be careful here is the link: https://gateoverflow.in/302802/gate2019-46

---

Noice  question

---

Choosing independently(Rpeatation is allowed ) so that,

#ways to select 2 numbers out of 7 numbers with MSB 0 = $7*7$

#ways to select 2 numbers out of 7 numbers with MSB 1 = $6*6$

total favourable ways = $7*7$  $+$ $6*6$ $= 78$

total ways $= 13*13=169$

probability = $78/169= 0.5029$

Answer 1

These are two groups:

1. MSB with $1$
2. MSB with $0$

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

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

$P = \frac{n_{\text{MSB1}}\;\ast\; n_{\text{MSB1}} + n_{\text{MSB0}}\;\ast \;n_{\text{MSB0}}}{n_{\text{Total}}}$

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

Comments

@Akash Papnai, @Satbir, @@akash.dinkar12, @srestha

1) why is the denominator  also 13^2 ? Is it because of independence of events ?

Is there any rule that ,for independence events, the Sample space will also be independent ?

 

2)Can someone pls elaborate the  use of the word “Uniformly” in this problem ?

 

Thanks in advance

---

@MohanK, 

1) Yes denominator is $13^2$, as both can be picked independently. You can think of it as picking up two cards from the deck of cards with replacement. So the second time when you will pick up the card, you will still have all the options open to choose from. Same goes here.

2) Uniformly at random means you have to pick one number randomly, and uniform means that the probability of picking a number is same across the set which will be $1/13$.

---

yes if the independent term not given then ans would have been :-0.462

$\frac{7C_{2} + 6C_{2}}{13C_{2}}$ =0.462

Answer 2

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.

Comments

same mistake :(  

the word "INDEPENDENT" plays a significant role here.

Answer 3

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

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

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

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

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

$\mathbf{NUMBER}$	$\mathbf{MSB}$			$\mathbf{LSB}$
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 $\mathbf{.5029}$

Comments

There should be only one 169 in denominator at the end

---

You are right but I am unable to edit it as it is showing some thing like “userid, content is needed”, something like that.
It seems there is some error with the editor as of now.

If someone have the edit authority please make the last line as

$\dfrac{49+36}{169}$

---

Now fully corrected.

Answer 4

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.

Comments

SILLY QUESTION ALERT:- Why denominator = 13*13? Is Independency the reason for this?

---

2 numbers are chosen out of 13

---

Got it Thnx :)