Case study problem probability 3 chapter 13 class 12

Case study problem probability 3 chapter 13 class 12

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Case study (Probability)

Read the following and answer the question(Case study problem probability 3 chapter 13 class 12)

Case study problem probability 3 chapter 13 class 12
There are two shops in a village market named Lila general store and Mina general store. In Lila general store 30 tin pure Mustard oil, 40 tin adulterated mustard oil while in Mina general store 50 tin pure mustard oil and 60 tin adulterated oil are there. Subhash babu wants to purchase one tin oil from any shop selecting at random.

(i) The conditional probability that an adulterated mustard oil tin is purchased at random, given that Lila general store is selected to purchase will be

(a) 8/7               (b) 4/7

(c) 5/7               (d) 2/7

(ii) The conditional probability that an adulterated mustard oil tin is purchased at random, given that Mina general store is selected to purchase will be

(a) 5/11             (b) 6/11

(c) 1/11             (d) 10/11

(iii) The total probability of purchasing adultrated mustard oil if shop[ and tin of oil are selected at random, is

(a) 40/77             (b) 33/77

(c) 7/33               (d) 43/77

(iv) Shubash babu wants to know quality of mustard oil. Before purchasing he selected first, a shop at random and then selected a tin of mustard oil at random. If the tin selected at random has adulterated oil, then the probability that the selected tin is of Lila general store is

(a) 22/43                 (b) 11/43

(c) 13/43                 (d) 7/13

(v) In above question (iv), the probability that the selected tin is of Mina general store is

(a) 22/43          (b) 11/43

(c) 21/43           (d) 7/43

Solution: Let X, Y and A be the event such that

X = Selection of Lila general store

Y = Selection of Mina general store

A = purchasing of a tin having adulterated mustard oil

P(X) = 1/2, P(Y) = 1/2

P(A/X) = 40/70, P(A/Y) = 60/110

(i) Answer (b)
P(An adulterated mustard oil tin is purchased at random, given that Lila general store)

P(A/X) = 40/70 = 4/7

(ii) Answer (b)

P( An adulterated mustard oil tin is purchased at random, given that Mina general store )

P(A/Y) = 60/110 = 6/11

(iii) Answer (d)

The total probability of purchasing adultrated mustard oil

P(A) = P(X)\times P(A/X) + P(Y)\times P(A/Y)

=\frac{1}{2}\times \frac{4}{7}+ \frac{1}{2}\times \frac{6}{11}

= \frac{2}{7}+\frac{3}{11}

= \frac{22+21}{77}=\frac{43}{77}

(iv) Answer (a)

P(the tin selected at random has adulterated oil, then the probability that the selected tin is of Lila general store)

P(X/A)= \frac{P(X)\times P(\frac{A}{X})}{P(X)\times P(\frac{A}{X})+P(Y)\times P(\frac{A}{Y})}

= \frac{\frac{1}{2}\times \frac{4}{7}}{\frac{1}{2}\times \frac{4}{7}+ \frac{1}{2}\times \frac{6}{11}}

= \frac{\frac{2}{7}}{\frac{2}{7}+\frac{3}{11}}

= \frac{2}{7}\times \frac{77}{43} = \frac{22}{43}

(v) Answer (c)

P(the tin selected at random has adulterated oil, then the probability that the selected tin is of Mina general store)

P(Y/A)= \frac{P(Y)\times P(\frac{A}{Y})}{P(X)\times P(\frac{A}{X})+P(Y)\times P(\frac{A}{Y})}

= \frac{\frac{1}{2}\times \frac{6}{11}}{\frac{1}{2}\times \frac{4}{7}+ \frac{1}{2}\times \frac{6}{11}}

= \frac{\frac{3}{11}}{\frac{2}{7}+\frac{3}{11}}

= \frac{3}{11}\times \frac{77}{43} = \frac{21}{43}

Some other case study question

https://gmath.in/case-study-problem-probability-1-chapter-13-class-12/

https://gmath.in/case-study-problem-probability-2-chapter-13-class-12/

Case study problem probability

Case study problem probability 1 chapter 13 class 12

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