In a test of a printed circuit board using a random test pattern, an array of 14 bits is equally likely to be 0 or 1. Assume the bits are independent. (a) What is the probability that all bits are 1s? Round your answer to six decimal places (e. g. 98.765432).

The probability that all 14 bits are 1s = 0.000061

Step-by-step explanation:

Probability that a bit is 1 is p = 0.5

Probability that a bit isn't 1 is q = 1 - 0.5 = 0.5

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 14

x = Number of successes required = 14

p = probability of success = 0.5

q = probability of failure = 0.5

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)¹⁴⁻¹⁴

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)⁰

P(X = 14) = 1 (0.000061)(1) = 0.000061

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation:

answer: the answer is b. i think hope i got it correct

step-by-step explanation: