# Probability

Question 1 |

8 / (2e ^{3}) | |

9 / (2e ^{3}) | |

17 / (2e ^{3}) | |

26 / (2e ^{3}) |

**Probability**

**Discuss it**

PR(X < 3) = Pr(x = 0) + Pr(x = 1) + Pr(x = 2) = f(0, 3) + f(1, 3) + f(2, 3) Put = 3 and k = 0, 1, 2 in the formula given at http://en.wikipedia.org/wiki/Poisson_distribution#Definition = 17 / (2e

^{3})

Question 2 |

13/90 | |

12/90 | |

78/90 | |

77/90 |

**GATE CS 2013**

**Probability**

**50 Aptitude & Reasoning MCQs with Answers**

**Discuss it**

Question 3 |

10/21 | |

5/12 | |

2/3 | |

1/6 |

**GATE CS 2012**

**Probability**

**Discuss it**

Solution set: { 6, (1,5), (1,6) ......}

i.e. P(6 appeared on first throw) +

P(1 appeared on first throw and 5 appeared on second throw) +

P(1 appeared on first throw and 6 appeared on second throw) + ....................

= 1/6 + (1/6)(1/6) + (1/6)(1/6) + .....

= 1/6 + 9/36

= 5/12.**Viewpoint 2:**

P(......) = P(6 came on first throw) + P(sum>= 6 and 1,2,3 appeared in first throw)

= 1/6 + ????

P(1,2,3 appeared in first throw) = 1/2 //P(E1)

P(sum >= 6 | 1,2,3 appeared in first throw) = 9/18 //P(E2 | E1)

// Our new sample space is: { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6) }

// 9 favorable cases: {(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) }

P(sum>= 6 and 1,2,3 appeared in first throw) = (1/2)(9/18) //P(E2 âˆ© E1) = P(E1)P(E2|E1)

P(what we are looking for) = 1/6 + 9/36 = 5/12

Correct Answer: B

Question 4 |

0 and 0.5 | |

0 and 1 | |

0.5 and 1 | |

0.25 and 0.75 |

**GATE CS 2012**

**Probability**

**Discuss it**

Question 5 |

1/3 | |

1/4 | |

1/2 | |

2/3 |

**GATE CS 2011**

**Probability**

**Discuss it**

Question 6 |

R = 0 | |

R < 0 | |

R >= 0 | |

R > 0 |

**GATE CS 2011**

**Probability**

**50 Aptitude & Reasoning MCQs with Answers**

**Discuss it**

**V**

**ariance**Â measures how far a set of numbers is spread out. (A variance of zero indicates that all the values are identical.) A non-zero variance is always positive:

Question 7 |

*X*= {

*x*

_{1},

*x*

_{2},...,

*x*

_{n}}. Let

*Î¼*

_{x}be the mean and

*Ïƒ*

_{x}be the standard deviation of

*X*. Let another finite sequence

*Y*of equal length be derived from this as

*y*

_{i}

*=*

*a*x*

_{i}

*+*

*b*, where

*a*and

*b*are positive constants. Let

*Î¼*

_{y}be the mean and

*Ïƒ*

_{y}be the standard deviation of this sequence. Which one of the following statements is INCORRECT?

Index position of mode of X in X is the same as the index position of mode of Y in Y. | |

Index position of median of X in X is the same as the index position of median of Y in Y. | |

Î¼_{y} = aÎ¼_{x}+b | |

Ïƒ_{y} = aÏƒ_{x}+b |

**GATE CS 2011**

**Probability**

**Discuss it**

Question 8 |

1/5 | |

4/25 | |

1/4 | |

2/5 |

**GATE CS 2011**

**Probability**

**50 Aptitude & Reasoning MCQs with Answers**

**Discuss it**

Question 9 |

pq + (1 - p)(1 - q) | |

(1 - q) p | |

(1 - p) q | |

pq |

**GATE CS 2010**

**Probability**

**Discuss it**

A computer can be declared faulty in two cases 1) It is actually faulty and correctly declared so (p*q) 2) Not faulty and incorrectly declared (1-p)*(1-q).

Question 10 |

^{99}is a multiple of 10

^{96}?

1/625 | |

4/625 | |

12/625 | |

16/625 |

**GATE CS 2010**

**Probability**

**Discuss it**

^{96}which are also divisor of 10

^{99}10

^{96}, 2x10

^{96}, 4x10

^{96}, 5x10

^{96}, 8x10

^{96}, 10x10

^{96}, 20x10

^{96}, 25x10

^{96}, 40x10

^{96}, 50x10

^{96}, 100x10

^{96}, 125x10

^{96}, 200x10

^{96}, 250x10

^{96}, 500x10

^{96}, 1000x10

^{96}The total number of divisors of 10

^{99Â }=Â 10000 Â (See http://www.math.cmu.edu/~mlavrov/arml/13-14/number-theory-09-29-13.pdf) So the probability = 16/10000 = 1/625