Case study problem probability 5 chapter 13 class 12

 Case study chapter 13 ( Probability)

Mahindra tractors is India’s leading farm equipment manufacturer. It is the largest tractor selling factory in the world.

Case study problem probability 5 chapter 13 class 12
Mahindra tractors is India’s leading farm equipment manufacturer. It is the largest tractor selling factory

This factory has two machines A and B. Past recordshows that machine A produced 60% and machine B produced 40% of the output(tractors). Further 2% of the tractors produced by machine A and 1% produced by machine B were defective. All the tractors are put into one big store hall and one tractor is chosen at random.(Case study problem probability 5 chapter 13 class 12)

Based on the above information answer the following question :

(i) Find the total probability of chosen tractor ( at random) is defective.

(ii)(a) If in random choosing, chosen tractor is defective, then find the probability that the chosen tractor is produced by machine ‘A’.

(b) If in random choosing, chosen tractor is defective, then find the probability that the chosen tractor is produced by machine ‘B’

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

X = Production of tractors by Machine A.

Y = Production of tractors by Machine B.

A = Choosing of defective tractors

P(X) = 60/100 = 3/5, P(Y) = 40/100 = 2/5

P(A/X) = 2/100 = 1/50, P(A/Y) = 1/100

(i) The total probability of chosen tractor ( at random) is defective

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

=\frac{3}{5}\times \frac{1}{50}+ \frac{2}{5}\times \frac{1}{100}

= \frac{3}{250}+\frac{1}{250}

= \frac{4}{250}=\frac{2}{125}

(ii) (a) P( chosen tractor is defective, then find the probability that the chosen tractor is produced by machine ‘A’)

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{3}{5}\times \frac{1}{50}}{\frac{3}{5}\times \frac{1}{50}+ \frac{2}{5}\times \frac{1}{100}}

= \frac{\frac{3}{250}}{\frac{3}{250}+\frac{1}{250}}

= \frac{3}{250}\times \frac{250}{4}

= \frac{3}{4}

(ii) (b) P(chosen tractor is defective, then find the probability that the chosen tractor is produced by machine ‘B’)

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{2}{5}\times \frac{1}{100}}{\frac{3}{5}\times \frac{1}{50}+ \frac{2}{5}\times \frac{1}{100}}

= \frac{\frac{1}{250}}{\frac{4}{250}}

= \frac{1}{4}

Question :A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots

Case study problem probability 6 chapter 13 class 12
A coach is training 3 players. He observes that the player A can hit a target 4

from this situation answer the following(Case study problem probability 6 chapter 13 class 12)

(a) Let A: the target is hit by A, B: the target is hit by B and C: The target is hit by C. Then find the probability that A, B and C all will hit

(b) Reffering to (a),what is the probability that B, C will hit and A will lose ?

(c) With reference to the events mentioned in (a), what is the probability that any two of A, B and C will hit ?

(d) What is the probability that none of them will hit the target ?

(e) What is the probability that at least one of A, B or C will hit the target ?

Solution : For solution please click here

 

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