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irina [24]
3 years ago
10

Among 14 electrical components exactly 3 are known not to function properly. If 6 components are randomly selected, find the fol

lowing probabilities: 1) The probability that all selected components function properly. 2) The probability that exactly 2 are defective. 3) The probability that at least 1 component is defective.a) (i) 0.1923 (ii) 0.4406 (iii) 0.3671.
b) (i) 0.0699 (ii) 0.4406 (iii) 0.3671.
c) (i) 0.0699 (ii) 0.4406 (iii) 0.9301.
d) (i) 0.0699 (ii) 0.3671 (iii) 0.9301.
e) (i) 0.1923 (ii) 0.0699 (iii) 0.3671.
f) none of the above.
Mathematics
1 answer:
mart [117]3 years ago
3 0

Answer:

1) Probability(All components function properly) = 2/13 = 0.1538

2) P(Exactly 2 are defective) = 30/91 = 0.3297

3) P(X≥1) = 0.8461

<u>f) none of the above</u>

Step-by-step explanation:

Total electrical components = 14

Components not functioning properly = 3

Components functioning properly = 14-3 = 11

Components selected = 6

1) If all selected components are functioning properly, that means all the 6 components were selected from the 11 components which were functioning properly so,

No. of ways to select components functioning properly = ¹¹C₆

Total no. of ways to select 6 components out of 14 = ¹⁴C₆

Probability(All components function properly) = No. of ways to select components functioning properly/Total no. of ways to select 6 components out of 14

                                                                            = ¹¹C₆ / ¹⁴C₆

                                                                            = 462/3003

Probability(All components function properly) = 2/13 = 0.1538

2) P(Exactly 2 are defective) can be calculated by considering that 2 are selected out of the 3 defective components and rest of the 4 components are selected from the pool of 11 components which are not defective.

P(Exactly 2 are defective) = ³C₂ ¹¹C₄ / ¹⁴C₆

                                           = (3 x 330)/3003

P(Exactly 2 are defective) = 30/91 = 0.3297

3) Now we need to calculate the probability P(X≥1). The simple way to calculate this probability is:

P(X≥1) = 1 - P(X<1)

           = 1 - P(X=0)

P(X=0) means no components are defective i.e. all components function properly. We have calculated this probability in part 1 so,

P(X≥1) = 1 - (2/13)

P(X≥1) = 0.8461

Since the answers to parts 1,2 and 3 do not match any of the options given here, the correct option is <u>f) none of the above.</u>

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