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Let S be the sample space.

Then, n(S) = numbers between 1 to 25 = 25

Let E = EVENT that the number selected is a prime number

⇒ E = {2, 3, 5, 7, 11, 13, 17, 19, 23}

n(E) = 9

∴ Probability of occurrence of event;

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

$(= {\rm{\;}}\frac{{10!}}{{5! \times 5!\;}}\; \times \frac{{5! \times 10!}}{{15!}} = \frac{{10! \times 10!}}{{15! \times 5!}})$

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

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

= 12/(11×13)

GIVEN that Number of BOYS = 20

Number of GIRLS = 30

Girls Getting A grade = 15 and boys getting A grade = 10

So, Probability of choosing a girl = 30/50 = 3/5

Probability of choosing a A grade student = 25/50 = 1/2 

Now A grade student can be a girl

So, probability of choosing it = 15/50 = 3/10 

So, REQUIRED probability of choosing a girl or a A grade student = (3/5 + 1/2) - 3/10 = 4/5

Quantity A:

Required probability = (⁴C₂ × ⁶C₂ × ³C₁)/¹³C₅

⇒ (6 × 15 × 3)/{(13 × 12 × 11 × 10 × 9)/120}

⇒ 30/143

⇒ 0.2098

Quantity B:

Required probability = (⁴C₁ × ⁶C₃ × ³C₁)/¹³C₅

⇒ (4 × 20 × 3)/{(13 × 12 × 11 × 10 × 9)/120}

⇒ 80/429

⇒ 0.1865

QUANTITY A:

Total number of outcomes = ⁷C₅ = 21

Since the REQUIRED probability is that the COMMITTEE has at least two ENGLISH teachers, there are two combinations:

2 maths teachers and 3 English teachers or 3 maths teachers and 2 English teachers;

∴ Number of FAVOURABLE outcomes = ⁴C₂ × ³C₃ + ⁴C₃ × ³C₂ = 6 + 12 = 18

∴ Probability that the committee has at least two English teachers = 18/21 = 6/7

Quantity B:

= 0.5

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

Odds against Sushant is 5 : 2

Probability that he will SOLVE question = 2/7

Odds in FAVOUR of Sushant is 7 : 5

Probability that he will solve question = 7/12

Problem will be solved means EITHER Sushant or Rahul or both will be able to solve

⇒ P (A or B) = P(A) + P(B) - P (A and B)

⇒ P (A or B) = 2/7 + 7/12 - (2/7 x 7/12)

⇒ P (A or B) = 2/7 + 7/12 - 1/6

⇒ P (A or B) = (24 + 49 - 14)/84

The probability that a heads occurs when this coin is tossed is 0.25.

⇒ Probability of occurrence of TAILS = 1 – 0.25 = 0.75

For exactly two tails to occur in 4 tosses, remaining two outcomes should be heads.

In 4 tosses, 2 tails can occur in ⁴C₂ WAYS, i.e., 6 ways.

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($(\bar B)$) = 1 – 1/2 = 1/2 and P($(\bar G)$) = 1 – 5/9 = 4/9

Probability that ONE of them will be selected = P(B) × P($(\bar G)$) + P($(\bar B)$) × P(G)

⇒ 1/2 × 4/9 + 1/2 × 5/9

⇒ 4/18 + 5/18 = 1/2