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Komok [63]
3 years ago
12

how many different license plates are possible if each contains 2 letters(out of the alphabet’s 26 letters) followed by 5 digits

(from 0 to 9)? how many of these license plates contain no repeated letters and no repeated digits?
Mathematics
1 answer:
timofeeve [1]3 years ago
3 0

Answer:

total number of license plates : 676000

total number of license plates with no repeated letters and no repeated digits : 468000

Step-by-step explanation:

i) the license plate is to be made up of two letters (out of 26 alphabets) followed by 5 digits ( from 0 to 9).

therefore the total number of combinations for license plates when repetitions of digits and numbers are allowed will be given

26 × 26 × 10 × 10 × 10 = 676000

ii) the number of license plates containing no repeated letters and no repeated digits will be given by

26 × 25 × 10 × 9 × 8 = 468000

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In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
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Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

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Let P(D) = Probability of a defective product

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P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

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P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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