CMAT Probability, Combinatorics Questions

There are 6 red balls, 11 yellow balls and 5 pink balls

Total number of balls = 6 + 11 + 5 = 22

Probability of selecting red ball = $$\frac{6}{22}$$

After selecting one ball, there are 21 balls left

Probability of selecting yellow ball = $$\frac{11}{21}$$

Therefore, probability = $$\frac{6}{22}\times\ \frac{11}{21}$$ = $$\frac{1}{7}$$

Answer is option C.

Number of ways of selecting 2 red balls and 1 green ball = $$15_{C_2}.\ 10_{C_1}$$

Number of ways of selecting 3 balls = $$25_{C_3}$$

Probability of selecting 2 red balls and 1 green ball = $$\ \frac{15_{C_2}.10_{C_1}}{25_{C_3}}$$ = $$\frac{21}{46}$$

Answer is option A.

Now we have 5 men and 3 women
committee of 4 to be made such that it has at least 1 woman
Case 1 :
1W and 3M
We get $$^5C_3\times\ ^3C_{1\ }=\ 30$$
Case 2 : 2W 2M
$$^5C_2\times\ ^3C_{2\ }=\ 30$$
Case 3 : 3W 1 M
$$^5C_1\times\ ^3C_{3\ }=\ 5$$
So total cases = 65.
so 1 is true
Now
ARRANGE
If vowels to be at even position :
we get cases as $$\frac{3!}{2!}\times\ \frac{4!}{2!}$$ = 36
So statement 2 is true

we have : 2435753228 
Now here we have 5 odd digits and 5 even digits and total 10 places 
so 5,5,3,3,7 will arrange in 5 places and 2,2,2,8,4 will arrange in 5 places 
we get : total arrangements =$$\frac{5!}{3!}\times\ \frac{5!}{2!\times\ 2!}$$
we get  $$\frac{\left(5!\right)^2}{3!\ \left(2!\right)^2}$$

Let the sample size of the above events be 6 units.

Since $$P(A)=\frac{2}{3}, P(B)=\frac{1}{2}$$ and $$P(A\cap B)=\frac{1}{3}$$,

Then event A will have 4 units($$\frac{2}{3}\times\ 6$$), event B will have 3 units and event $$A\cap B$$ will have 2 units.

This can be expressed in terms of the Venn diagram as follows:

Now, it can be observed that $$A\cap B^c\ =2$$ units.

Thus, $$P\left(A\cap B^c\right)\ =\frac{2}{6}=\frac{1}{3}$$.

Similarly, $$A\ U\ B^c\ =5$$ units($$A=4$$ and $$B^c-A=1$$ units)

So, $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$

Since only option A has $$P\left(A\cap B^c\right)=\frac{1}{3}$$ and $$P\left(A\ U\ B^c\right)\ =\frac{5}{6}$$,

The correct option is A.

Probability that A releases a mobile = 0.7
Probability that mobile is a success given that it is released by A = 0.8
Probability that mobile is released by A and is success = 0.8*0.7 = 0.56
And probability that mobile is a success and released by B = 0.28
Total probability for success = 0.56+0.28 = 0.84
Now Probability that it is success and is released by A = 0.56/0.84 = 2/3

Number of ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women

= $$C^{10}_6\times C^8_5$$

= $$C^{10}_4\times C^8_3$$

= $$\frac{10\times9\times8\times7}{1\times2\times3\times4}\times\frac{8\times7\times6}{1\times2\times3}$$

= $$(10\times3\times7)\times(8\times7)$$

= $$210\times56=11760$$

=> Ans - (D)

Probability of brother's selection = $$\frac{1}{8}$$ and probability of brother's rejection = $$\frac{7}{8}$$

Probability of sister's selection = $$\frac{1}{5}$$ and probability of sister's rejection = $$\frac{4}{5}$$

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)

= $$(\frac{1}{8})(\frac{4}{5})+(\frac{7}{8})(\frac{1}{5})$$

= $$\frac{4}{40}+\frac{7}{40}=\frac{11}{40}$$

=> Ans - (A)

There are 2 physics teacher.They can stand in a row in $$P(2,2)=2!=2\times1=2$$ ways

Similarly, the two Chemistry and Mathematics teacher can stand in 2 ways each.

These three sets can be arranged in themselves in $$3!=3\times2\times1=6$$ ways

=> Required number of ways = $$2\times2\times2\times6=48$$

=> Ans - (B)

Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters.

These letters can be arranged in $$4! = 24$$ ways

The letters EA may be arranged amongst themselves in $$2!=2$$ ways.

Number of arrangements having vowels together = $$(24 \times 2) = 48$$ ways

Total arrangements of all letters = $$5!=120$$ ways

=> Number of arrangements not having vowels together = $$(120 - 48) = 72$$ ways

=> Ans - (D)

Based on the given constraints, there will be two cases in which task 1 is either assigned to one among person 3 or 4 or to one among person 5 or 6

Case 1:

Assigning Task 1 to either Person 3 or 4.

No. of ways of assigning task 1 = 2 ways

No. of ways of assigning task 2 = 1 way 

Remaining 4 tasks can be assigned among the remaining 4 persons in 4! ways.

Therefore, Total number of ways = 2 x 1 x 4! = 48

Case 2:

Assigning Task 2 to either Person 5 or 6.

No. of ways of assigning task 1 = 2 ways

No. of ways of assigning task 2 = 2 ways 

Remaining 4 tasks can be assigned among the remaining 4 persons in 4! ways.

Therefore, Total number of ways = 2 x 2 x 4! = 96

Hence, total number of possible arrangements= 48 + 96 = 144

Probability of A or B = Probability of A + Probability of B - Probability of A and B.

$$\Rightarrow$$ Probability of drawing king or heart = Probability of drawing king + Probability of drawing heart - Probability of drawing kings which are hearts from the deck of cards.

A deck of cards contain 52 cards, of which there are 4 kings and 13 hearts and 1 king which is heart also. 

Probability of drawing a king = $$\frac{4}{52}$$

Probability of drawing a heart = $$\frac{13}{52}$$

Probability of drawing a king which are heart also = $$\frac{1}{52}$$

$$\Rightarrow$$ Probability of drawing king or heart = $$\frac{4}{52} + \frac{13}{52} - \frac{1}{52}$$

= $$\frac{16}{52}$$

= $$\frac{4}{13}$$

When a pair of dice are thrown simultaneously the maximum sum possible is 12.

The possible cases of getting a sum of 9 are:

                    dice 1     dice 2

                    four         five.....................(1)

                    five          four.....................(2)

                     six          three....................(3)

                   three         six.......................(4)  

So a total of 4 cases are possible

The possible cases of getting a sum of 10 are:

                    dice 1     dice 2

                     four        six......................(1)

                      six         four.....................(2)

                     five        five.....................(3)

So a total of 3 cases are possible

The possible cases of getting a sum of 11 are:

                   dice 1       dice 2

                    five           six...................(1)

                     six           five..................(2)

So a total of 2 cases are possible

The possible cases of getting a sum of 12 are:

                  dice 1      dice 2

                   six           six.....................(1)

So a total of 1 case is possible.

Total number ways of getting a sum of 9 or more = 4 + 3 + 2 + 1 = 10 ways

Total outcomes when two dice are rolled simultaneously = 6 * 6 = 36 ways

Probability of getting a sum of 9 or more = Expected number of outcomes/ Total outcomes

= $$\frac{10}{36}$$

= $$\frac{5}{18}$$

Probability = Expected number of outcomes/ Total number of outcomes.

Total number of outcomes =  The number of ways of selecting 3 professors at random from a total of 10 professors.

= 10C3 = 120 ways.

Expected number of outcomes = Selecting 3 professors so that at least one of the junior professor is selected.

The possible cases are,

Case(1): Selecting 1 junior professor out of 4 and Selecting 2 senior professor out of 6 = 4C1 * 6C2.

Case(2): Selecting 2 junior professor out of 4 and Selecting 1 senior professor out of 6 = 4C2 * 6C1.

Case (3): Selecting 3 junior professor out of 4  = 4C3.

So Expected number of outcomes = 4C1 * 6C2 + 4C2 * 6C1 + 4C3 = $$4\times 15 + 6\times 6 + 4$$ = 100 ways.

Probability = $$\frac{100}{120} = \frac{5}{6}$$ 

Probability = Expected number of outcomes/ Total number of outcomes.

Total number of outcomes we get in a single throw of two dice = $$6\times 6$$ = 36.

Possible cases of getting 'nine' in a single throw of two dice:

                  dice 1           dice 2

                  three             six..................(1)

                  six                three................(2)

                  four              five..................(3)

                  five              four...................(4)

So, total of 4 cases.

Possible cases of getting 'ten' in a single throw of two dice:

                dice 1            dice 2

                four                six.....................(1)

                 six                 four....................(2)

                five                 five....................(3)

So, total of 3 cases.

Expected number of outcomes = Total possible cases of getting 'nine' or 'ten'  in a single throw of two dice = 4 + 3 = 7 .

So, Probability = $$\frac {7} {36}$$

Probability = $$Expected  number  of  outcomes\div Total  number  of  outcomes$$.

Total number of outcomes = Total number of ways to draw randomly 3 balls out of 12 balls = 12C3 ways.

Expected number of outcomes = Number of ways to draw 2 yellow balls out of 8 and 1 black ball out of 4 = $$8C2\times 4C1$$

$$ Hence Probability = 8C2\times 4C1\div 12C3 = 28\times 4\div 220 = \frac{28}{55}$$

Three coins are tossed in the air and two of the coins land with tails face upwards.

Assuming that the coins are fair and the first part does not try to indicate that they are not fair. The outcomes that make at least two coins heads are : 

HHT, HHH, HTH, THH 

Thus, outcome is 4 out of 8 = 50%

=> Ans - (A)

Since balls of same color should come together, let us consider 3 red balls as one unit and 2 blue balls as one unit and 4 yellow balls as another unit.

So, we have a total of 3 different units which can be arranged in '3!' ways.

these 3 red balls can internally be arranged in '3!' ways.

Similarly the blue balls and yellow balls can be arranged internally in '2!' and '4!' ways respectively.

So, total number ways = $$3!\times 3!\times 2!\times 4!= 6\times 6\times 2\times 24$$= 1728 ways