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Question: A batch of fifty radio sets was purchased from 3 different companies A, B and C. Eighteen of them we...

A batch of fifty radio sets was purchased from 3 different companies A, B and C. Eighteen of them were manufactured by A, twenty of them by B and the rest were manufactured by C. The companies A and C. The companies of A and C produce excellent quality radio sets with probability equal to 0.9. B produces the same with probability 0.6. What is the probability of t3he event that the excellent quality radio set chosen at random is manufactured by the company B?

Explanation

Solution

Hint : The probability ( to produce excellent quality sets) and total number of sets manufactured by each company are given and we are supposed to find the probability of excellent quality radio set chosen at random is manufactured by the company B.
This can be done by calculating favorable outcomes for B with respect to the total favorable outcomes.
Probability = Number of favorable outcomesTotal outcomes{\text{Probability = }}\dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Total outcomes}}}}

Complete step-by-step answer :
We have been given:
Total items manufactured by:
A= 18
B= 20
C= 50-(18+20) [ As the rest are manufactured by C)
=12
Probability to produce excellent radio sets:
P(A) = 0.9
P(B) = 0.6
P(C) = 0.9
The favorable outcomes for each company can be given as:
Probability = Favorable outcomesTotal outcomes{\text{Probability = }}\dfrac{{{\text{Favorable outcomes}}}}{{{\text{Total outcomes}}}}
Favorable outcomes = Probability × Total outcomes{\text{Favorable outcomes = Probability }} \times {\text{ Total outcomes}}
For A: For B: For C:
P (A). A P (B). B P(C).C
Calculating the probability that excellent quality radio set chosen at random is manufactured by the company B:
Probability of B = Favorable outcomes for BTotal Favorable outcomes (sum of favorable outcomes of each company) {\text{Probability of B = }}\dfrac{{{\text{Favorable outcomes for B}}}}{{{\text{Total Favorable outcomes (sum of favorable outcomes of each company) }}}}
P(B)=P(B).BP(A).A+P(B).B+P(C).CP(B) = \dfrac{{P(B).B}}{{P(A).A + P(B).B + P(C).C}}
Substituting values, we get:
P(B)=0.6×200.9×18+0.6×20+0.9×12P(B) = \dfrac{{0.6 \times 20}}{{0.9 \times 18 + 0.6 \times 20 + 0.9 \times 12}}
P(B)=1216.2+12+10.8P(B) = \dfrac{{12}}{{16.2 + 12 + 10.8}}
P(B)=1239P(B) = \dfrac{{12}}{{39}}
P(B)=413 or 0.3P(B) = \dfrac{4}{{13}}{\text{ or }}0.3
Therefore, the probability that the excellent quality radio set chosen at random is manufactured by company B is 413 or 0.3\dfrac{4}{{13}}{\text{ or }}0.3

Note : In such questions, we keep on noting the things given in question line by line, then apply the formulas according to what needs to be found to get the desired results. One must correctly apply the formula for total outcome where we must consider the probability of all the events.