CBSE Board Examination 2023 · Section E · 4 marks

Q36: Case study: three types of Yoga in a Venn diagram

Board 2023Ch 13 · Probability

The 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 :

Types of Yoga: Hatha Yoga, Bikram Yoga, Vinyasa Yoga, Kundalini Yoga, Anusara Yoga.

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 a member performing type C Yoga is 0.44.

The Venn diagram: A only 0.32, A and B only 0.09, B only y, B and C only x, C only 0.21, none of them 0.11.

On the basis of the above information, answer the following questions :

(i) Find the value of x. (1 mark)

(ii) Find the value of y. (1 mark)

(iii) (a) Find $P\left(\dfrac{C}{B}\right)$ ⁠. (2 marks)

(iii) (b) Find the probability that a randomly selected person of the society does Yoga of type A or B but not C. (2 marks)

Exam answer

(i) $P(C) = x + 0.21 = 0.44$

Why?

$P(C)$ is the total of all the regions inside oval C. From the diagram these are $x$ (B and C only) and 0.21 (C only). Oval C does not meet oval A.

$\Rightarrow x = 0.23$

Why?

$x = 0.44 - 0.21 = 0.23$ ⁠.

\begin{aligned} x + 0.21 &= 0.44 \\ x &= 0.44 - 0.21 \\ &= 0.23 \end{aligned}

Watch out: 0.44 is the whole of C, not the region marked "C only".

(ii) The probabilities in S add up to 1:

\begin{aligned} &0.32 + 0.09 + y + 0.23 \\ &\quad + 0.21 + 0.11 = 1 \\ \Rightarrow\ &0.41 + y + 0.44 + 0.11 = 1 \end{aligned}

Why?

The six regions of the diagram do not overlap, and together they make up the whole sample space S. So their probabilities add up to 1.

In the second line the terms are grouped: $0.32 + 0.09 = 0.41$ ⁠, and $0.23 + 0.21 = 0.44$ ⁠, which is $P(C)$ ⁠.

$\Rightarrow y = 0.04$

Why?

$0.41 + 0.44 + 0.11 = 0.96$ ⁠, so $y = 1 - 0.96 = 0.04$ ⁠.

\begin{aligned} 0.41 + 0.44 &= 0.85 \\ 0.85 + 0.11 &= 0.96 \\[4pt] y + 0.96 &= 1 \\ y &= 1 - 0.96 = 0.04 \end{aligned}

(iii) (a)

P\left(\frac{C}{B}\right) = \frac{P(C \cap B)}{P(B)}

Why?

This is the definition of conditional probability. B is the event known to have happened, so $P(B)$ goes in the denominator.

\begin{aligned} P(B) &= 0.09 + 0.04 + 0.23 \\ &= 0.36 \\ P(C \cap B) &= x = 0.23 \end{aligned}

Why?

$P(B)$ adds the three regions inside oval B: 0.09 (A and B only), $y = 0.04$ (B only) and $x = 0.23$ (B and C only).

\begin{aligned} 0.09 + 0.04 &= 0.13 \\ 0.13 + 0.23 &= 0.36 \end{aligned}

$C \cap B$ is the part shared by B and C, which is the region $x$ ⁠.

$\therefore\ P\left(\frac{C}{B}\right) = \frac{0.23}{0.36} = \frac{23}{36}$

Why?

Multiply the top and the bottom by 100 to clear the decimals:

\begin{aligned} \frac{0.23}{0.36} &= \frac{0.23 \times 100}{0.36 \times 100} \\ &= \frac{23}{36} \end{aligned}

23 is prime and does not divide 36, so the fraction is in lowest terms.

(iii) (b)

\begin{aligned} &P(\text{A or B but not C}) \\ &= 0.32 + 0.09 + 0.04 \end{aligned}

Why?

"A or B" means the regions inside oval A or oval B. "But not C" removes any region inside oval C.

That leaves A only (0.32), A and B only (0.09) and B only ( $y = 0.04$ ⁠). The region $x$ is left out, because it lies inside C.

= 0.45

Why?

$0.32 + 0.09 = 0.41$ ⁠, and $0.41 + 0.04 = 0.45$ ⁠.

Check another way: $P(A \cup B)$ is $0.32 + 0.09 + 0.04 + 0.23 = 0.68$ ⁠. Taking away the part inside C, $x = 0.23$ ⁠, leaves 0.45.

$\therefore$ Required probability $= 0.45$ ⁠.

Where this comes from

Chapter 13, Probability: Venn diagram; conditional probability.

Question type: (iii)(a) Given numbers

How to spot it: click to see

(iii)(a) Type 1. Given numbers: divide by the event after the bar

Spot it: the question gives $P(E)$ ⁠, $P(F)$ and $P(E\cap F)$ ⁠, and asks for $P(E|F)$ or $P(F|E)$ ⁠.

Here: The Venn diagram gives the numbers: $P(C \cap B) = x = 0.23$ and $P(B) = 0.09 + 0.04 + 0.23 = 0.36$ ⁠. $B$ is after the bar, so divide by 0.36.

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