Probability

Correct Answer – Option 3 : 0.7627

Concept: 

If there can be only 2 cases of true and false, then:

Probability of r true cases out of n (≥ r) P(x = r) = ⁿC_r p^rq^(n-r) 

where p is the probability of case being true and q is probability of case being false

Note: p and q are ≤ 1

Calculation:

The probability of correct answer p = \(\rm 1\over 4\) = 0.25

The probability of wrong answer q = \(\rm 3\over 4\) = 0.75

Total questions n = 5

Probability of atleast one correct answer (x ≥ 1):

P(x ≥ 1) = 1 – P(x = 0)

⇒ P(x ≥ 1) = 1 – ⁵C₀ p⁰q⁵

⇒ P(x ≥ 1) = 1 – (0.75)⁵ 

⇒ P(x ≥ 1) = 1 – 0.2373

⇒ P(x ≥ 1) \(\boldsymbol{\rm \approx}\) 0.7627

Correct Answer – Option 2 : 1 and 3 only

Concept:

Let U be the universal set and A, B, C be the subsets of U.

If \(\rm A∩ B ∩ C = \phi\) then  A, B, C are mutually exclusive.

Çalculations:

Given, U = {1, 2, 3, …., 20}.

Let A, B, C be the subsets of U.

A be the set of all numbers which are perfect squares

⇒ A = {1, 4, 9 16}

B be the set of all numbers which are multiples of 5

⇒ B = {5, 10, 15, 20}

and C be the set of all numbers, which are divisible by 2 and 3

⇒ C = {6, 12, 18}

Now, \(\rm A∩ B ∩ C = \phi\)

So, Å, B, C are mutually exclusive.

Hence statement 1 is correct

Here  Å, B, C are mutually exclusive so A, B, C can’t be mutually exhaustive

Hence statement 2 is wrong

A ∪ B = {1, 4, 5, 9, 10, 15, 16, 20}

n(A ∪ B) = 8

U = {1, 2, 3, …., 20}

n(U) = 20

Now, The number of elements in the complement set of A ∪ B  = n(U) – n(A ∪ B) = 20 – 8 = 12

Hence statement 3 is correct

Correct Answer – Option 3 : Quantity I ˃ Quantity II

Quantity I:

Total number of outcomes = ¹⁶C₄

⇒ 16!/(4! × 12!)

⇒ (16 × 15 × 14 × 13 × 12!) / (4 × 3 × 2 × 1 × 12!)

⇒ 1820

Number of favourable outcomes = ⁵C₂ × ³C₂

⇒ 30

⇒ Required probability = 30/1820

⇒ 0.0165 (approx)

Quantity II:

Total number of possible ways to drawn a card = ⁵²C₅

⇒ 52!/(5! × 47!)

⇒ (52 × 51 × 50 × 49 × 48 × 47!)/ (5 × 4 × 3 × 2 × 1 × 47!)

⇒ 2598960

Number of sets of 5 with all are jack = ⁴C₄ × ⁴⁸C₁

⇒ 1 × (48 × 47!) / (1! × 47!) = 48

⇒ P(E ) = 48/2598960

⇒ 1/54145

⇒ 0.000018(approx)

∴ Quantity I > Quantity II

Correct Answer – Option 3 : 63

Calculation:

Let the number of green balls and blue balls present in the bag be ‘g’ and ‘b’ respectively

Also, number of red balls = 17

Total number of balls = (17 + g + b)

Probability of drawing a red ball = 17/(17 + g + b)

Probability of drawing a green ball = g/(17 + g + b)

Probability of drawing a blue ball = b/(17 + g + b)

Now a/q we have

⇒ g/(17 + g + b) = 1/7 + 17/(17 + g + b)

⇒ (g – 17)/(17 + g + b) = 1/7 

⇒ 6g – b = 136    …..(i)

also, 

⇒ b/(17 + g + b) = 1/21 + 17/(17 + g + b) 

⇒ (b – 17)/(17 + g + b) = 1/21

⇒ 20b – g = 374    …..(ii)    

using equation (i) and (ii), we get 

g = 26, b = 20

∴ Required total number of balls in the bag = (26 + 20 + 17) = 63