Advance Math Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts

ADVANCE MATH PRACTICE TEST [2 - EXERCISES]

Question : 1

Suppose six coins are tossed simultaneously. Then the probability of getting at least one tail is :

a) \${53}/{54}\$

b) \${63}/{64}\$

c) \${71}/{72}\$

d) \$1/{12}\$

Answer: (b)

If six coins are tossed, then the total no. of outcomes = \$(2)^6\$ = 64

Now, probability of getting no tail = \$1/{64}\$

Probability of getting at least one tail = 1 - \$1/{64} = {63}/{64}\$

Question : 2

Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2 ?

a) 81

b) 91

c) 36

d) 116

Answer: (b)

There can be 3 cases :

I. When one dice shows 2.

II. When two dice shows 2.

III. When three dices shows 2.

Case I : The dice which shows 2 can be selected out of the 3 dices in \$^3C_1\$ ways.

Remaining 2 dices can have any 5 numbers except 2. So number of ways for them = \$^5C_1\$ each, so no of ways when one dice shows 2 = \$^3C_1 × ^5C_1 × ^5C_1\$ .

Case II : Two dices, showing 2 can be selected out of the 3 dices in \$^3C_2\$ ways and the rest one can have any 5 numbers except 2, so number of ways for the remaining 1 dice = 5.

So, number of ways, when two dices show

2 = \$^3C_2\$ × 5

Case III : When three dices show 2 then these can be selected in \$^3C_3\$ ways.

So, number of ways, when three dices show

2 = \$^3C_3\$ = 1

As, either of these three cases are possible.

Hence, total number of ways

= (3 × 5 × 5) + (3 × 5) + 1 = 91

Question : 3

There are 6 different letters and 6 correspondingly addressed envelopes. If the letters are randomly put in the envelopes, what is the probability that exactly 5 letters go into the correctly addressed envelopes?

a) \$1/6\$

b) \$1/2\$

c) Zero

d) \$5/6\$

Answer: (c)

As there are 6 letters and envelopes, so if exactly 5 are into correctly addressed envelopes, then the remaining 1 will automatically be placed in the correctly addressed envelope. Thus, the probability that exactly 5 go into the correctly addressed envelope is zero.

Question : 4

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is \$1/2\$ , \$1/3\$ and \$1/4\$ . Probability that the problem is solved is

a) \$1/2\$

b) \$2/3\$

c) \$3/4\$

d) \$1/3\$

Answer: (c)

\$P(E_1) = 1/2, P(E_2) = 1/3 and P(E_3) = 1/4;\$

P\$(E_1 ∪ E_2 ∪ E_3)\$ = 1 - P(\$\ov{E_1}\$) P (\$\ov{E_2}\$) P (\$\ov{E_3}\$)

= 1 - \$(1 - 1/2)(1 - 1/3)(1 - 1/4) = 1 - 1/2 × 2/3 × 3/4 = 3/4\$

Question : 5

The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat and the terminal digits are even is

a) 72

b) 288

c) 144

d) 720

Answer: (d)

The first and the last (terminal) digits are even and there are three even digits. This arrangement can be done in \$^3P_2\$ ways. For any one of these arrangements, two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits (leaving out the terminal digits) may be arranged using these 5 digits in \$^5P_4\$ ways. The required number of numbers is \$^3P_2 × ^5P_4\$ = 6 × 120 = 720.

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