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

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