Advance Math Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

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.

Answer: (a)

Only one of A and B can be alive in the following, mutually exclusive ways.

$E_1$ A will die and B will live

$E_2$ B will die and A will live

So, required probability = $P(E_1) + P (E_2)$

= p(1 - q) + q (1 - p) = p + q - 2pq.

Answer: (b)

The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.

∴ No. of choices available to the student

= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.

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.

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

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 - $1/{64} = {63}/{64}$