We are given a random test pattern for a printed circuit board, represented by an array of 10 bits. Each bit is independent and is equally likely to be 0 or 1. This means each bit has:

$P(\text{bit} = 1) = \frac{1}{2}$ $P(\text{bit} = 0) = \frac{1}{2}$

(a) What is the probability that all bits are 1s?

For all bits to be 1, each of the 10 bits must independently be 1. Thus:

$P(\text{all bits are 1}) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

$P(\text{all bits are 1}) = 0.0009765625$

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(b) What is the probability that all bits are 0s?

This is the same as part (a), except all bits must be 0. The probability is the same:

$P(\text{all bits are 0}) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

$P(\text{all bits are 0}) = 0.0009765625$

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(c) What is the probability that exactly 5 bits are 1s and 5 bits are 0s?

This is a binomial probability problem, where we want exactly 5 bits to be 1 out of 10 bits. The binomial formula is:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ Where:

• $n = 10$ (number of bits)
• $k = 5$ (number of bits that should be 1)
• $p = \frac{1}{2}$ (probability of each bit being 1)

$P(X = 5) = \binom{10}{5} \cdot \left(\frac{1}{2}\right)^5 \cdot \left(\frac{1}{2}\right)^5 = \binom{10}{5} \cdot \left(\frac{1}{2}\right)^{10}$

$\binom{10}{5} = 252$ $P(X = 5) = 252 \cdot \frac{1}{1024}$ $P(X = 5) = \frac{252}{1024} \approx 0.2461$

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Final Answers

Part	Probability
(a) All 1s	0.0009765625
(b) All 0s	0.0009765625
(c) Exactly 5 ones and 5 zeros	0.2461

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Would you like me to explain any step in more detail or clarify anything?

5 Related Questions

1. What if the bits were not independent? How would the calculations change?
2. How does the probability change if the array length increases to 20 bits?
3. How do you calculate the expected number of 1s in a 10-bit array?
4. What is the probability of having at least 7 ones in the 10-bit array?
5. How would the calculation change if the probability of each bit being 1 was $0.6$ instead of $0.5$?

Tip

When solving binomial probability problems, always check whether the events (bits) are independent — this simplifies calculations significantly.