There are different types of Yoga which involve

Case Study:1

Read the following passage and answer the following question. There are different types of Yoga which involve the usage of different poses of Yoga Asanas, Meditation and Pranayam as shown in the figure below: The venn diagram below represents the probabilities of three different types of Yoga, A, B and C performed by the people of a society. Further, it is given that probability of member performing type C Yoga is 0.44

(i) Find the value of x.

(ii) Find the value of y.

iii)(a)Find P(C/B).

(b) Find the probability that a randomly selected person of the cociety does Yoga of the type A or B but not C.

Solution:

(i) Given that probability of a member performing type C type is 0.44.

∴ x + 0.21 = 0.44

⇒ x = 0.44 – 0.21 = 0.23

⇒ x = 0.23

(ii) 0.32 + 0.09 + y + x = 1 – 0.11

⇒ 0.32 + 0.09 + y + x = 0.89

⇒ 0.85 + y = 0.89

⇒ y = 0.89 – 0.85

⇒ y = 0.04

(iii) (a) $P(C/B) = \dfrac{P(C\cap B)}{P(B)}$

$= \dfrac{x}{0.09 + y + x } = \dfrac{0.23}{0.09 + 0.04 + 0.23}$

$= \dfrac{0.23}{0.36} = \dfrac{23}{36}$

(iii)(b) Required probability = 0.32 + 0.09 + y = 0.41 + y

= 0.41 + 0.04 = 0.45

Case study 2

Read the following passage and answer the following question. A shopkeeper sells three types of flower seed $A_1,A_2,A_3$ they are sold in the form of a mixture, where the proportion of these seeds are 4 : 4 : 2, respectively. The germination rates of the three types of seeds are 45%, 60%, and 35% respectively

(a) Calculate the probability that a randomly choosen seed will germinate.

(b) Calculate the probability that the seed is of the type $A_2$ , given that a randomly choosen seed germinates.

Solution:

$A_1$ = Event of choosing flower seed $A_1$

$A_2$ = Event of choosing flower seed $A_2$

$A_3$ = Event of choosing flower seed $A_3$

Let E = The event that seed germinates

$P(A_1) = \dfrac{4}{4+4+2} = \dfrac{4}{10}$

$P(A_2) = \dfrac{4}{4+4+2} = \dfrac{4}{10}$

$P(A_3) = \dfrac{2}{4+4+2} = \dfrac{2}{10}$

$P(E/A_1) = \dfrac{45}{100}$ $P(E/A_1) = \dfrac{60}{100}$

$P(E/A_1) = \dfrac{35}{100}$

(a) the probability that a randomly choosen seed will germinate

$P(E) = P(A_1).P(E/A_1) + P(A_2).P(E/A_2) + P(A_3).P(E/A_3)$

$= \dfrac{4}{10}\times \dfrac{45}{100} + \dfrac{4}{10}\times \dfrac{60}{100} + \dfrac{2}{10}\times \dfrac{35}{100}$

$= \dfrac{180}{1000} + \dfrac{240}{1000} + \dfrac{70}{1000}$

$= \dfrac{180 + 240 + 70}{1000}$ $= \dfrac{490}{1000} = 0.49$

(b) the probability that the seed is of the type $A_2$ , given that a randomly choosen seed germinates.

$P(A_2/E) = \dfrac{P(A_2).P(E/A_2)}{P(A_1).P(E/A_1) + P(A_2).P(E/A_2) + P(A_3).P(E/A_3)}$

$= \dfrac{\dfrac{4}{10}.\dfrac{60}{100}}{\dfrac{4}{10}\times \dfrac{45}{100} + \dfrac{4}{10}\times \dfrac{60}{100} + \dfrac{2}{10}\times \dfrac{35}{100}}$