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+17 votes

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 _______________.

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

+30 votes

Best answer

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.

i.e. If we select 2 and 1 then 2,1 and 1,2 will be treated as different selections.

+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.

+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.

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.

+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.

+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.

+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 $\mathbf{.5029}$

+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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