Question

In: Statistics and Probability

2. (a) The owner of a juice bar, Sam, orders oranges from 3 suppliers, namely A,...

2. (a) The owner of a juice bar, Sam, orders oranges from 3 suppliers, namely A, B and C. From past record, each box of oranges from suppliers A, B, C contains 1%, 3% and 10% rotten oranges respectively. Assume that Sam orders 70% of the oranges from supplier A, 15% of the oranges from supplier B and 15% of the oranges from supplier C. One morning, a box of oranges arrives but Sam does not know which supplier sent it. A random check is then conducted to see if the oranges are rotten.

i. Find the probability that a randomly selected orange from the box is rotten.

ii. Find the probability that there are more than 1 rotten orange among 5 randomly selected oranges.

iii. Sam inspects 10 oranges, and finds that 1 is rotten. With this information, find the probability that the box comes from supplier C.

Expert Solution

Let rotten oranges be denoted by 'R'

Probability that orange is rotten given it is supplied from A = P(R | A)

= 0.01 (1%)

Similarly P(R | B ) = 0.03

P(R | C) = 0.10

Probability of supplying from supplier A = P(A) = 0.70

Similarly P(B) = 0.15

P(C) = 0.15

i. Find the probability that a randomly selected orange from the box is rotten.

If it is rotten it can be from supplier A, B or C

P(Rotten) =

= P(R | A) P(A) + P(R | B) P(B) + P( R|C ) P(C)

= 0.01 * 0.70 + 0.03 * 0.15 + 0.10 * 0.15

ii. Find the probability that there are more than 1 rotten orange among 5 randomly selected oranges.

We can model this as a binomial distribution

Where X is the event of being a rotten orange and the probability is 0.0265

P(X =x) =

P(More than 1 is rotten) = P( X > 1)

= 1 - P( X 1 )

= 1 - [ P( X = 0) + P( X = 1) ]

= 1 - ()

= 1 - (0.874 + 0.119)

iii. Sam inspects 10 oranges, and finds that 1 is rotten. With this information, find the probability that the box comes from supplier C.

The probability of getting 1 rotten tomato out of 10 =

Where 10C1 = no. ways that 1 tomato is rotten eg. 1st, 2nd, ..10th

= probability that 1 is rotten

= probability that 9 are not rotten.

P( from supplier C | 1 rotten out of 10 ) =

Where the numerator considers probability from only supplier C and the denominator considers any supplier.

= 0.015

P( from supplier C | 1 rotten out of 10 ) =

orchestra answered 2 years ago

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