CMAT Probability and Combinatorics Questions

The probability will be $$\dfrac{1}{2}$$ because each coin toss is an independent event. The results of the first four tosses do not influence the outcome of the fifth toss, so the probability of getting a head on the fifth toss remains $$\dfrac{1}{2}$$.

The given number is 8314629. Arranging the digits in descending order, the number is 9864321.

We are to find pairs of digits such that the distance between them is same in both the numbers. 

On inspection we find only 4 such pairs, (9,4) , (9,6), (6,4), and (4,2) exist .

Hence, the correct option is FOUR. 

Each of the 4 people can have a birthday in any of the 12 calendar months. The total number of ways the birthdays can be distributed is = $$12\times\ 12\times\ 12\times\ 12=12^4$$

Now, for the birthdays to fall in exactly two calendar months, we must first choose which two months these will be. The number of ways of choosing = $$^{12}C_2=\dfrac{12\times\ 11}{2}=66$$

For each person, their birthday can be in one of the two chosen months. This gives $$2^4=16$$ total possibilities.

However, this includes two cases where all four birthdays fall in just one of the two months (either all in the first chosen month or all in the second). We must subtract these two cases to ensure the birthdays fall in exactly two months.

So, number of ways = $$2^4-2=16-2=14$$ possibilities.

Total number of favourable cases = $$66\times\ 14$$

So, required probability = $$\dfrac{66\times\ 14}{12\times\ 12\times\ 12\times\ 12}=\dfrac{77}{1728}$$

So, option A is the correct answer.

For the ninth one to be the last defective, among the first 8 watches, first 3 has to be defective.

So, total number of ways of selecting = $$^8C_3$$

And total number of possible cases = $$^{15}C_4$$

(Out of the 15 watches, examined in order, any 4 can be defective)

So, required probability = $$\dfrac{^8C_3}{^{15}C_4}=\dfrac{\dfrac{8!}{3!\times\ 5!}}{\dfrac{15!}{4!\times\ 11!}}=\dfrac{8}{195}$$

So, correct answer is option B.

1.) First and second in Mathematics:

There are 30 students eligible for the first prize and 29 students remaining for the second prize.

So, total number of ways of distributing the prize = $$30\times\ 29$$

2.) First and second in Physics:

There are 30 students eligible for the first prize and 29 students remaining for the second prize.

So, total number of ways of distributing the prize = $$30\times\ 29$$

3.) First in Chemistry:

There are 30 students eligible for the first prize.

So, total number of ways of distributing the prize = 30

4.) First in English:

There are 30 students eligible for the first prize.

So, total number of ways of distributing the prize = 30

So, total number of ways of giving all the prizes = $$\left(30\times\ 29\right)\times\ \left(30\times\ 29\right)\times\ 30\times\ 30=30^4\times\ 29^2$$.

So, correct answer is option D.

There are 20 girls in total, out of which two are A and B.

So, out of remaining 18 girls, we need to choose 4 girls so that they can sit between A and B.

Number of ways of choosing these 4 girls = $$^{18}C_4$$

Number of ways of arranging these 4 girls = $$4!$$

Number of ways of arranging the remaining 14 girls = $$14!$$

Also, number of ways in which $$A$$ and $$B$$ can arrange = $$2!$$

So, total number of cases such that there are 4 girls between A and B = $$^{18}C_4\times\ 4!\times\ 14!\times\ 2!$$

Also, number of ways of arranging 20 girls in round table = $$\left(20-1\right)!=19!$$

So, required probability = $$\dfrac{^{18}C_4\times\ 14!\times\ 4!\times\ 2!}{19!}=2\times\ \dfrac{18!}{19!}=\dfrac{2}{19}$$

The probability he hits the disc = Probability of hitting on any of the four shots

P = (Hitting of 1st) + (Not hitting on 1st shot*Hitting on 2nd shot) + (Not hitting on 1st and 2nd shot*Hitting on 3rd shot) + (Not hitting on 1st, 2nd and 3rd shot*Hitting on 4th shot)

We know that the Probability of not hitting at a shot = 1 - The probability of hitting the bottle at that shot

$$P = 0.1+\left(0.9\times0.2\right)+\left(0.9\times0.8\times0.35\right)+\left(0.9\times0.8\times0.65\times0.45\right)$$

$$P=0.1+0.18+0.252+0.2106$$

$$P=0.7426$$

Hence, the answer is Option C

There are 10 stations on the railway line. A journey ticket is for traveling between two different stations, where the order matters (i.e., traveling from station A to B is different from traveling from B to A).

For each starting station, there are 9 possible destinations (since you can't travel to the same station). Since there are 10 possible starting stations, the total number of different journey tickets is $$10\times9=90$$

"OMEGA" comprises 5 letters: O, M, E, G, A.

Vowels: O, E, A (3 vowels)

Consonants: M, G (2 consonants)

Odd positions in a 5-letter word are: 1st, 3rd, and 5th.

The vowels O, E, and A can be placed in these positions in 3! = 6 ways.

The even positions are 2 and 4.

The consonants M and G can be arranged in these positions in 2! = 2 ways.

The total number of arrangements is the product of the arrangements of vowels and consonants = 3! x 2! = 6*2 = 12 ways.

A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles.

If a single marble is chosen at random from the jar, the probability of getting a yellow marble =$$\frac{\text{no of yellow marbles}}{\text{Total no of mables}}$$= $$\frac{4}{10}$$

Probability of getting a green marble = $$\frac{\text{no of green marbles}}{\text{Total no of mables}}$$ = $$\frac{3}{10}$$

Probability of getting either a yellow or a green marble = $$\frac{\text{no of yellow or green marbles}}{\text{Total no of mables}}$$ = $$\frac{4+3}{10}$$ = $$\frac{7}{10}$$

Probability of getting either a red or a yellow marble = $$\frac{\text{no of yellow or red marbles}}{\text{Total no of mables}}$$ = $$\frac{4+1}{10}$$ = $$\frac{5}{10}$$

So, only B, C are correct.

A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles.

If a single marble is chosen at random from the jar, the probability of getting a yellow marble =$$\frac{\text{no of yellow marbles}}{\text{Total no of mables}}$$= $$\frac{4}{10}$$

Probability of getting a green marble = $$\frac{\text{no of green marbles}}{\text{Total no of mables}}$$ = $$\frac{3}{10}$$

Probability of getting either a yellow or a green marble = $$\frac{\text{no of yellow or green marbles}}{\text{Total no of mables}}$$ = $$\frac{4+3}{10}$$ = $$\frac{7}{10}$$

Probability of getting either a red or a yellow marble = $$\frac{\text{no of yellow or red marbles}}{\text{Total no of mables}}$$ = $$\frac{4+1}{10}$$ = $$\frac{5}{10}$$

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)