Question

1. Given an equilateral triangle ABC with a side of 5 cm. Find the probability that a point taken at random will be located from point A at a distance greater than 2 cm.

2. There are two boxes: inside the first 10 white and 15 black balls, inside the second 10 white and 10 black balls. 5 balls are transferred from the first box to the second and then one ball is removed from the second box at random, what is the probability that it is white?

Thanks Sir )

Answer 1

Solution

Back-up Theory

Probability of an event E, denoted by P(E) = n/N ………………………………………………(1)

where

n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Area of an equilateral triangle of edge a = {(√3)/4}a².......................................................(2)

Area of sector of angle θ° and radius r = (θ/360)πr²...................................................(3)

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}……(4)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN……. (4a)

Now, to work out the solution,

Q1

A point taken at random will be located from point A at a distance greater than 2 cm, if the point is within the triangle but outside the sector of 60° of radius 2cm at the vertex A.

So, vide (1), n = area outside the sector and N = area of the equilateral triangle.

Given equilateral triangle ABC has a side of 5 cm, vide (2), araea of the triangle = 25{(√3)/4} = 13.0833

Vide (3), area of sector of angle θ° and radius r = (θ/360)πr² = 25π/6 = 10.8253

Hence, the required probability

= (13.0833 - 10.8253)/13.0833

= 2.258013.0833

= 0.1726 Answer 1

Q2

When 5 balls are transferred from the first box to the second, the second box has 25 balls now. The number of white balls in the second box now would be:

10 if no white ball is transferred, i.e., 0 white and 5 balck are transferred, for which probability is

{¹⁰C₀ x¹⁵C₅)/ ²⁵C₅ [(4) and (1)]

= 3003/53130 [vide (4a)]

= 0.0562

11 if exactly one white ball and 4 balck are transferred, for which probability is

{¹⁰C₁ x¹⁵C₄)/ ²⁵C₅ [(4) and (1)]

= 13650/53130 [vide (4a)]

= 0.2569

12 if exactly 2 white balls and 3 balck are transferred, for which probability is

{¹⁰C₂ x¹⁵C₃)/ ²⁵C₅ [(4) and (1)]

= 20475/53130 [vide (4a)]

= 0.3584

13 if exactly 3 white balls and 2 balck are transferred, for which probability is

{¹⁰C₃ x¹⁵C₂)/ ²⁵C₅ [(4) and (1)]

= 21600/53130 [vide (4a)]

= 0.0889

14 if exactly 4 white balls and 1 balck are transferred, for which probability is

{¹⁰C₄ x¹⁵C₁)/ ²⁵C₅ [(4) and (1)]

= 3150/53130 [vide (4a)]

= 0.0127

14 if exactly 5 white balls and 0 balck are transferred, for which probability is

{¹⁰C₅ x¹⁵C₀)/ ²⁵C₅ [(4) and (1)]

= 240/53130 [vide (4a)]

= 0.0008.

Thus, the required probability

= {0.0562 x (10/25) +   {0.2569 x (11/25) + {0.3584 x (12/25) + {0.0889 x (13/25) + {0.0127 x (14/25)

+ {0.0008 x (15/25)

= 0.3744 Answer

DONE