Probability Concepts for CAT

Ans: 

If 0 < x < 45° ,sin x < cos x

 x = 45°, sin x = cos x

If 45°<x≤90°,sinx>cosx

If 90° < x ≤ 180°, sin x > cos x (as sin x is always + ve

In that zone)

If 180°<x<225°,sinx>cosx

x=225°,sinx=cosx

If 225°<x<270°,sinx<cosx

Probability = \(\frac{(90-45+ (180-90)+ (225-180)}{270}= \frac{180}{270} = \frac{2}{3}\)

Ans: (0.1

There are 6 vertices in a hexagon. Using 3 vertices out of 6 vertices we can form ⁶C₃ triangles. But there can be only triangles out of ⁶C₃ triangles which are equilateral

(i) ∆ACE (ii) ∆BDF.

Required Probability = (2/⁶C₃) =(2/20) = 0.1

Ans: (C)

Let A be the event getting exactly 3 defectives in the examination of 8 pens and B be the event of getting nineth pen defective.

P(A) = (⁴C₃ x ¹¹C₅)/¹⁵C₈

P (\(\frac{B}{A}\)) = 1/7

Required probability = P(A ∩ B) = P(A)P(\(\frac{B}{A}\)) = \(\frac{8}{195}\)

Ans: (D)

probability of at least on red ball = 1 − (Probability of all blue balls)

probability of drawing one blue ball is (1/2)

probability of drawing all four blue balls will be (1/2)⁴ = (1/16)

probability of at least on red ball = 1- (1/16) = 15/16

Ans: (B)

There are seven levels above the main floor.

 7⁵ is the total number of ways each of the five people can exit the cabin on any of the seven floors.

And the optimal number of ways, i.e. the number of ways in which the five individuals evacuate the building via different floors, is ⁷P₅.

Required Probability = (⁷P₅/7⁵) = 0.15

Ans: (C)

As the red balls are smallest of all the type of balls.

If we select one out all the red balls then

Required probability = 1/6

Ans: (B)

E₁ = The event in which A speaks truth
 

E₂ = The event in which B speaks truth

P(E₁) = \(\frac{100}{60}= \frac{3}{5}\)      P(E₁’) = 1 – \(\frac{3}{5}= \frac{2}{5}\)

P(E₂) = \(\frac{80}{100}= \frac{4}{5}\)      P(E₂’) = 1 – \(\frac{4}{5}= \frac{1}{5}\)

Required probability = P[(E₁ ∩ E₂)∪ (E’₁ ∩ E’₂)] =P(E₁ ∩E₂)+P(E’₁ ∩E’₂) = P(E₁). P(E₂)+ P(E₁). P(E₂)

 = \((\frac{3}{5} \times \frac{4}{5}) + (\frac{2}{5} \times \frac{1}{5}) = \frac{14}{25}= 0.56\)

Ans: (C)

There are (64 x 63) distinct methods to select two distinct squares. Each of the four corner squares has a favorable case count of two. The favorable number of cases for each of the 24 non-corner squares on all four sides of the chessboard is 3. For each of the remaining 36 squares, the number of favorable circumstances is 4.

the total number of favourable cases = 4 x2 + 24 x 3 + 36 x 4 = 224

Required probability = \(\frac{224}{64 \times 63}= \frac{1}{18}\)

Ans: (A)

The last two digits can be dialled in 10 P2 = 90 ways.

Out of these 90 cases only one case is favourable.

So, the required probability =

Ans: (B)

3 squares on a chessboard have ⁶⁴C₃ possible selections. There are two mutually exclusive ways to select two squares of one color and a third square of a distinct color.

• 2 white and one black
• 2 black and one white

No of favourable cases = ³²C₂ × ³²C₁ + ³²C₁ × ³²C₂ =2(³²C₂ × ³²C₁) = 1056

Total outcomes = ⁶⁴C₃= 41664

Required probability = (1056/41664) = (16/21)

Ans: (A)

Total number of words that can be formed = \(\frac{11!}{4!4!2!}\)

When all the S’s are together then the number of words = \(\frac{8!}{4!2!}\)

Required Probability = \(\frac{\frac{11!}{4!4!2!}}{\frac{8!}{4!2!}} = \frac{4}{165}\)

Ans: (C)

A be the event of getting exactly 3 defectives in the examination of 8 wristwatches.

B be the event of getting ninth wristwatch defective.

P(A) = (⁴C₃ x ¹¹C₅)/¹⁵C₈

P\((\frac{B}{A})\) = 1/7

Required probability = P(A ∩ B) = P(A)P\((\frac{B}{A})= \frac{8}{195}\)

Ans: (B)

Let x and y be the two non-negative integers since x + y = 200

(xy)_max = 100 x 100 = 10000

​

Favourable no of ways = 15-50+1 = 101

Total number of ways = 200

Probability = 100/200

Ans: (B)

Let E_i(i = 1, 2, 3 etc.) denote the event of drawing an even

numbered card in i^th draw and F_i (i = 1, 2, 3) denote the

event of drawing an odd numbered card in i^th draw, then

required probability

=P[(E₁ ∩F₂ ∩F₃)∪(F₁ ∩E₂ ∩F₃) ∪(F₁ ∩F₂ ∩E₃)]

=P(E₁) P(F₂) P(F₃) + P(F₁) P(E₂) P(F₃) + P(F₁) P(F₂) P(E₃)

= \(\frac{4}{9} \times \frac{5}{9} \times \frac{5}{9} + \frac{5}{9} \times \frac{4}{9} \times \frac{5}{9} + \frac{5}{9} \times \frac{5}{9} \times \frac{4}{9}= \frac{100}{243}\)

Ans: (A)

A = The first number is less than the second number

B = The third number lies between the first and the second

Let the selected numbers be x₁, x₂, x₃. Then they satisfy exactly one of the following inequalities.

x₁ <x₂ <x₃,  x₁ <x₃ <x₂,  x₂ <x₁ <x₃,  x₂ <x₃ <x₁,  x₃ <x₁ <x₂,  x₃ <x₂ <x₁

The total number of ways of selecting three numbers and then arranging them = ⁿC₃× 3! = ⁿP₃

P(A) = (ⁿC₃× 3)/( ⁿC₃× 3!)

P(B) = ⁿC₃/( ⁿC₃× 3!)

P\((\frac{B}{A})= \) P(A ∩ B)/ P(A) = 1/3

Ans: (C)

Since b and c each can assume 9 values from 1 to 9.

So, total number of ways of choosing band cis 9 × 9 = 81 Now, x² + bx + c > 0 for all x ∈ R

the possible values of b and c for which b² < 4c

required probability = 32/ 81

Ans: (B)

ASSISTANT → AA I N SSS TT, STATISTICS → A II C SSS TTT Here N and C are not common and same letters can be A, I, S, T.

Probability of choosing A = (²C₁/⁹C₁) x (¹C₁ /¹⁰C₁) = (1/45)

Probability of choosing I = (1/⁹C₁) x (²C₁ /¹⁰C₁) = (1/45)

Probability of choosing S = (³C₁/⁹C₁) x (³C₁ /¹⁰C₁) = (1/10)

Probability of choosing T = (²C₁/⁹C₁) x (³C₁ /¹⁰C₁) = (1/15)

Required Probability = (1/45)+ (1/45)+ (1/10)+ (1/15) = (19/90)