**Question 1:**

A coin of diameter 'd' is tossed which lands on a square tile of side 'a'. What is the probability that the coin will fall completely within the tile

[1] $\frac { ( a - d ) ^ { 2 } } { a ^ { 2 } }$

[2] $\frac { d ( 2 a - d ) } { a ^ { 2 } }$

[3] $\frac { d ( 2 a - d ) } { ( a - d ) ^ { 2 } }$

[4] $\frac { ( a - d ) ^ { 2 } } { d ( 2 a - d ) }$

**Answer & Solution**

Sample space = Area of the tile $= \mathrm { a } ^ { 2 }$

The favourable area is the inner square

of area $( a - d ) ^ { 2 }$

Therefore, required probability $= \frac { ( a - d ) ^ { 2 } } { a ^ { 2 } }$

**Question 2:**

In a plane, S lines of lengths 2, 3, 4, 5 and 6 cm are lying. What is the probability that by joining the three randomly chosen lines end to end a triangle cannot be formed?

[1] $\frac { 3 } { 10 }$

[2] $\frac { 7 } { 10 }$

[3] $\frac { 1 } { 2 }$

[4] 1

**Answer & Solution**

We know that the sum of two sides in a triangle is greater than the third side. Therefore, in following 3 cases, the triangle is not formed.

2. 3, 5

2, 3, 6

2, 4, 6

Total number of triangles formed using 5 lines $= 5 \mathrm { C } _ { 3 }$

Therefore, required probability $=\frac{3}{{}^{5}{{C}_{3}}}=\frac{3}{10}$

**Question 3:**

There are 7 boys and 8 girls in a class. A teacher has 3 items viz a pen, a pencil and an eraser, each 5 in number. He distributes the items, one to each student. What is the probability that a boy selected at random has either a pencil or an eraser?

[1] 2/3

[2] 2/21

[3] 14/45

[4] None of these

**Answer & Solution**

The probability of a boy being selected $= 7 / 15$. Therefore, the probability of having either a pencil or an eraser with him $= \frac { 7 } { 15 } \times \frac { 10 } { 15 } = \frac { 14 } { 45}$

**Question 4:**

5 army men are standing in a row left to right. Chetan and Chetak are charioteers. Shailendra and Surendra are the soldiers while Dinesh is a doctor. Dinesh always stands in the middle. The two soldiers and two charioteers do not stand next to each other respectively. If Shailendra is not standing immediately next to the doctor, then out of all the arrangements possible, what is the probability that a charioteer stands next to the doctor?

[1] 1/4

[2] 1/2

[3] 1/3

[4] 1

**Answer & Solution**

Let Sh -— Shailendra, Su — Surendra be soldiers, 'Ch -— Chetan, Ck – Chetak be charioteers, D — Dinesh — Doctor. Sh at only extreme places (Hence Sh — 1 or 5 ) and since Su can’t stand atleast just before Sh, this place has to be taken by a charioteer. Hence a charioteer always stands next to a doctor. Hence, P = 1.

**Question 5:**

There are 3 concentric circular strips on a dart. The Probability of hitting the inner most circular lamina is 1/9, that of the central strip is 1/3 and of the outer most strip is 5/9. One gets 10 points for hitting the inner most lemma, 6 for the central strip and 2 for the outermost strip. What is the probability of getting atleast 20 points in 3 attempts? Given that a target (i.e. the one of the 3 strips) is never missed?

[1] $\frac { 1 } { 81 }$

[2] $\frac { 5 } { 81 }$

[3] $\frac { 52 } { 729 }$

[4] $\frac { 1 } { 9 }$

**Answer & Solution**

20 points or more can be obtained in the following hits:

No. of Throws | Points in Throws | Probability | Net Probability | ||

1 | 2 | 3 | |||

1 | 10 | 10 | 10 | $\frac { 1 } { 9 } \times \frac { 1 } { 9 } \times \frac { 1 } { 9 }$ | $\frac { 1 } { 9 ^ { 3 } }$ |

2 | 10 | 10 | 6 | $\frac { 1 } { 9 } \times \frac { 1 } { 9 } \times \frac { 1 } { 3 }$ each | $3 \times \left( \frac { 1 } { 9 } \times \frac { 1 } { 9 } \times \frac { 1 } { 3 } \right)$ |

10 | 6 | 10 | |||

6 | 10 | 10 | |||

3 | 10 | 10 | 2 | $\frac { 1 } { 9 } \times \frac { 1 } { 9 } \times \frac { 5 } { 9 }$ each | $3 \times \left( \frac { 1 } { 9 } \times \frac { 1 } { 9 } \times \frac { 5 } { 9 } \right)$ |

10 | 2 | 10 | |||

2 | 10 | 10 | |||

4 | 10 | 6 | 6 | $\frac { 1 } { 9 } \times \frac { 3 } { 9 } \times \frac { 3 } { 9 }$ each | $3 \times \left( \frac { 1 } { 9 } \times \frac { 3 } { 9 } \times \frac { 3 } { 9 } \right)$ |

6 | 10 | 6 | |||

6 | 6 | 10 |

Therefore, Required probability $= \frac { 1 } { 9 ^ { 3 } } + \frac { 9 } { 9 ^ { 3 } } + \frac { 15 } { 9 ^ { 3 } } + \frac { 27 } { 9 ^ { 3 } } = \frac { 52 } { 9 ^ { 3 } } = \frac { 52 } { 729 }$

**Question 6:**

In the previous question, if hitting any of the 3 strips had equal probability, then what is the probability of the event in the previous question to occur

[1] $\frac { 1 } { 3 }$

[2] $\frac { 4 } { 9 }$

[3] $\frac { 10 } { 27 }$

[4] $\frac { 25 } { 81 }$

**Answer & Solution**

There are 3 strips with

10 | 6 | 2 |

points in the 1^{st} hit, one could hit either of the 3 strips,

Similarly in the 2^{nd} and 3^{rd} attempts one of the strips will be hit.

Hence there are 3 x 3 x 3 = 27 possibilities of hitting the target in 3 attempts.

Therefore, required probability=10/27

**Question 7:**

A locker at the RBI building can be opened by dialling a fixed three digit code (between 000 and 999). Chhota Chetan, a terrorist, only knows that the number is a three digit number and has only one six. Using this information he tries to open the locker by dialling three digits at random. The probability that he succeeds in his endeavor is

[1] $\frac { 1 } { 243 }$

[2] $\frac { 1 } { 900 }$

[3] $\frac { 1 } { 1000 }$

[4] $\frac { 1 } { 216 }$

**Answer & Solution**

Chhota Chetan knows that the code is a three digit number. Total number of numbers that can be formed = 3x(9x9x1) = 243.

Of these only one number will open the locker.

The probability of his success = 1/243.

**Question 8:**

A pair of fair dice are rolled together, till a sum of either 5 or 7 is obtained. The probability that the sum 5 happens before sum 7 is

[1] 0.45

[2] 0.4

[3] 0.5

[4] 0.6

**Answer & Solution**

5 can be obtained by getting any of [(4, 1), (3, 2). (2. 3). (1. 4)]

7 can be obtained by getting any of [(6, 1). (5, 2). (4. 3). (3. 4). (2. ‘5). (1. 6)]

Probability that 5 comes before 7 = 4/10 = 0.4

**Question 9:**

In the previous question, what is the probability of getting sum 7 before sum 5?

[1] 0.6

[2] 0.55

[3] 0.4

[4] 0.5

**Answer & Solution**

Probability of getting 7 before $5 = \frac { 6 } { 4 + 6 } = 0.6 = \frac { 6 } { 10 } = \frac { 3 } { 5 }$

**Question 10:**

Numbers are selected at random one at a time, from the numbers 00, 01, 02,…., 99 with replacement. An event E occurs if and only if the product of the two digits of a selected number is 18. If four numbers are selected, then the probability that E occurs at least 3 times is

[1] $\frac { 97 } { 390625 }$

[2] $\frac { 98 } { 390625 }$

[3] $\frac { 97 } { 380625 }$

[4] $\frac { 97 } { 380626 }$

**Answer & Solution**

Out of numbers 00, 01, 02,…., 99, those numbers the product of whose digits is 18 are 29,36, 63, 92 i.e., only 4.

$p = P ( E ) = \frac { 4 } { 100 } = \frac { 1 } { 25 } , q = P ( E ) = 1 - \frac { 1 } { 25 } = \frac { 24 } { 25 }$

Let X be the random variable, showing the number of times E occurs in 4 selections. Then p(so that E occurs at least 3 times) = p(X = 3 or X = 4)

$= p ( x = 3 ) + p ( x = 4 ) = 4 C _ { 3 } p ^ { 3 } q ^ { 1 } + 4 C _ { 4 } p ^ { 4 } q ^ { 0 } = 4 p ^ { 3 } q + p ^ { 4 }$

$= 4 \times \left( \frac { 1 } { 25 } \right) ^ { 3 } \times \frac { 24 } { 25 } + \left( \frac { 1 } { 25 } \right) ^ { 4 } = \frac { 97 } { 390625 }$