Total possible outcomes, n(S) = 6

Odd numbers in the range are 1, 3, 5

∴ Total count of odd numbers, n(E) = 3

P(E) = n(E)/(n(S) = 3/6 = 1/2

∴ The probability of getting an odd number = 1/2 

Number of ways of outcomes when two dice are thrown = n(S) = 36 and the possible cases of event when the sum of numbers on two dice is a prime number are

(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5).

number of events = 15

Probability = 15/36 = 5/12

Probability of A for not getting selected in interview = 1 – (1/5) = 4/5

Probability of B for not getting selected in interview = 1 – (1/6) = 5/6

So the required probability = $\frac{4}{5}\times \frac{5}{6}=\frac{2}{3}$

∵ Probability = [No of Favorable Outcomes /No. of Total Outcomes]

⇒ No. of Total Outcomes = 2⁶ = 64

⇒ No. of Favorable Outcomes = getting 3 heads out of 6 tossed $=6_{C_3}=\frac{6!}{3!\times 3!}=20$

⇒ Probability = 20/64

= 5 / 16

One-third of 15 mangoes got rotten

Rotten mangoes = 15/3 = 5

So, number of good mangoes = 15 – 5 = 10

When 5 mangoes are taken out randomly, the probability that no mango is rotten = ¹⁰C₅ /¹⁵C₅

$=\frac{10!}{5!\times 5!}\times \frac{5!\times 10!}{15!}=\frac{10!\times 10!}{15!\times 5!}$

$=\frac{10\times 9\times 8\times 7\times 6\times 5!\times 10!}{15\times 14\times 13\times 12\times 11\times 10!\times 5!}$

$=\frac{10\times 9\times 8\times 7\times 6}{15\times 14\times 13\times 12\times 11}$

= 12/(11×13)

Let S be the sample space.

Then, total number of balls = (4 + 6 + 8 + 10) = 28 balls

Thus, n(S) = 28

Let E = event that the ball drawn is black

∵ There are 10 black balls in the bag

Thus, n(E) = 10

∴ Probability of occurrence of event;

Maximum number of cases = ⁵⁰C₅ [because we pick any 5 numbers and they can be arranged to form x1,x2 … x5 in the given order]

If X₃ = 30, then X₁ and X₂ are less than 30 i.e. from 1 to 29 and X₄ and X₅ are greater than 30 i.e. from 31 to 50.

Number of ways of picking X₁ and X₂ out of 29 numbers = ²⁹C₂

Number of ways of picking X₄ and X₅ out of 20 numbers = ²⁰C₂

∴ Total number of ways of selecting X₁, X₂, X₃, X₄ and X₅ such that X₃ = 30 is ²⁹C₂ × ²⁰C₂

On tossing three coins, n(S) = 2 × 2 × 2 = 8

Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (∵ H = head, T = tail)

Number of events with at least two heads, n(E) = 4

On throwing two dice, n(S) = 6 × 6 = 36

Number of events of getting a total of 7, n(E) = 6, i.e. (1,6), (2,5) , (3,4), (4,3) , (5,2), (6,1)

Total no. of fruits = 7

No. of fruits to be drawn = 3.

∴ Total no. of ways of doing it = ⁷C₃

For drawing out the same fruits, either all of them must be apples or bananas.

∴ No. of possible ways of drawing all 3 apples = ³C₃

Also no. of possible ways of drawing 3 out of 4 bananas = ⁴C₃

∴ Total no. of possible ways of drawing out same fruit = ³C₃ + ⁴C₃

Probability that all the fruits drawn are same = (³C₃ + ⁴C₃)/⁷C₃

Total number of caps in box = (2 blue + 4 red + 5 green + 1 yellow) caps = 12

Let,

E1 = event of picking a blue cap

E2 = event of picking a yellow cap

Then,

P(E1 ) = 2/12

P(E2) = 1/12

From statement I:

Largest number of dice is a multiple of 6.

⇒ Numbers on dice can be from 19 to 24, or they can be 25 to 30. In both cases, the probability that a multiple of 5 is the outcome when the dice is thrown is different.

∴ Statement I alone is not sufficient to answer the question.

From statement II:

Smallest number of dice is not a multiple of 5. To have two multiples of 5 on dice, both smallest and largest numbers on dice must be multiples of 5.

⇒ In this case, there would be exactly one multiple of 5 on the dice. Probability will be 1/6.

∴ Statement II alone is sufficient to answer the question.

From statement I:

The word is not of four letters. The word can be of any number of letters except 4, from 1 to infinity. So, the probability will be 1/(1 + 1 + 1 +        ….infinite times), that will be 0.

∴ Statement I alone is sufficient to answer the question.

From statement II:

The word is not of two letters. The word can be of any number of letters except 4, from 1 to infinity. So, the probability will be 1/(1 + 1 + 1 +      ….infinite times), that will be 0.

∴ Statement II alone is sufficient to answer the question.

Quantity A:

2 digits numbers less than 50 which are divisible by 2 or 3 or 5 but not by 6 and 15

= 10, 14, 16, 20, 21, 22, 25, 26, 27, 28, 32, 33, 34, 35, 38, 39, 40, 44, 46

⇒ n(E) = 19

n(S) = Total number of 2 digits numbers less than 50 (10 to 49)

⇒ n(S) = 40

⇒ P(E) = n(E)/n(S) = 19/40

Quantity B:

Probability of selection of a boy in a interview is 1/2 and that of a girl is 5/9.

⇒ P(B) = 1/2 and P(G) = 5/9

⇒ P($\overline{B}$) = 1 – 1/2 = 1/2 and P($\overline{G}$) = 1 – 5/9 = 4/9

Probability that one of them will be selected = P(B) × P($\overline{G}$) + P($\overline{B}$) × P(G)

⇒ 1/2 × 4/9 + 1/2 × 5/9

⇒ 4/18 + 5/18 = 1/2

Sample space for the given random experiment to throw a die is, S = {1,2,3,4,5,6}

⇒ n(S) = 6

First we will find P(A),

P(A): Probability of getting a prime number.

Let A be the event of getting a prime number.

⇒ A = {2, 3, 5}

⇒ n(A) = 3

∴ P(A) = n(A)/n(S) = 3/6 = 1/2

Now,

P(B): Probability of getting a number lying between 2 and 6.

Let B be the event of getting a number lying between 2 and 6.

⇒ B = {3, 4, 5}

⇒ n(B) = 3

∴ P(B) = n(B)/n(S) = 3/6 = 1/2

∴ Quantity A = Quantity B