Aptitude

1

I forgot the last digit of a 7 digit telephone number. If one randomly dial the final three digits after correctly dialling the four, then what is the chance of dialling the correct number?

a

1/1000

b

1/1001

c

1/999

d

1/990

It is given that last three digits are randomly dialled. Then each of the digit can be selected out of 10 digits in 10 ways.

Hence required probability

= 1/10 * 1/10 * 1/10

= 1/1000

2

A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

a

A gain of Rs 27

b

A loss of 37

c

Neither gain or a loss

d

A gain or a loss depending upon the order in which the wins and losses occur

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

=64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs 27

Hence the final result is: 64-27 =37

3

A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?

a

7/108

b

5/54

c

1/9

d

2/17

The only way to roll higher on one die is if the magicians rolls between 2 and 5, inclusive, with two dice. Where he to roll a six with two dice than there is no way he could eclipse that number by rolling one die.

Below is the probability of rolling a certain number with two dice.

‘2’ – 1/36

‘3’ – 2/36

‘4’ – 3/36

‘5’-  4/36

Now the only numbers a magician can roll with the one die and win is between 3 and 6, inclusive. The chances of rolling any are always 1/6.

Next, we have to combine the probability distribution relating to two dice with that relating to the one.

The chances of a magician rolling any given number with one die are 1/6. So let’s start with the lowest number he can roll: a ‘3.’ To win with this roll, he will have to roll a ‘2’ with two dice, the odds of which are 1/36. So .

So the chances of him rolling a ‘3’ with one die and winning are 1/216.

Let’s repeat this logic for the next roll, ‘4.’ Chances of rolling are 1/6. Only way a ‘4’ wins is if he rolls a ‘2’ or a ‘3’ with two dice.

Odds of rolling a ‘2’ with two dice  + odds of rolling a ‘3’ with two dice = . Combine this with the odds of rolling a ‘4’ ( which is 1/6): .

If he rolls a ‘5’ with one die, he can win if he rolls a ‘4’ with two dice, the probability is 3/36. He can also win if he rolls a ‘2’ or a ‘3’ with two dice, the number outcomes we just found: 3/36. So we add

Next, if he rolls a ‘6’ on one die, he can beat ‘2’ through ‘5’ with the two dice. Number of ways to roll a ‘5’ = 4/36. Combining this with the odds of rolling a ‘6’ on one die with the odds or rolling ‘2’, ‘3’, ‘4’, or ‘5’ with the two dice we get: 10/216.

You may be wondering why I left the denominator as 216. Well, this allows us to add up all the instances he can possibly win:

4

In a single throw of two dice , find the probability that neither a doublet nor a total of 8 will appear.

a

7/15

b

5/18

c

13/18

d

3/16

n(S) = 36

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) }

Required probability = 10/36 = 5/18

Note: (4,4) occurs in both. Count once.

5

A division of a company consists of seven men and five women.  If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?

a

1/6

b

2/5

c

2/9

d

5/33

First, the denominator. We have twelve different people, and we want a combination of two selected from these twelve.  We will use the formula :

nC2 = (n(n-1))/2

which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers.  Here

12C2 = (12*(11))/2 = 6 * 11 = 66

That’s the total number of pairs we could pick from the twelve employees.  That’s our denominator.

For the numerator, we want every combination of two from the five female employees.  That’s

5C2 = (5*(4))/2 = 5 * 2 = 10

That’s the  number of pairs of female employees we could pick from the five.  That’s our numerator.

probability = 10/66 = 5/33

6

There are 10 items in a box, out of which 3 are defective. 2 items are taken one after the other. What is the probability that both of them are defective?

a

4/60

b

3/60

c

2/60

d

Both A and B

probability = 3C2/10C2 = ((3!) / (2! * 1!))/((10!) / (8! * 2!)) = 1/15
1/15 * 4/4 = 4/60