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+4 votes

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

+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

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

+6 votes

Best answer

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

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

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

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 $\mathbf{.5029}$

+2 votes

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

So, $((7*7) + (6 * 6) )/ (13 * 13)$

which is 0.5029

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