Ask question

Ask question

All categories

2 years ago

10

Suppose that 1% of (binary) bits received by your smartphone have errors. What is the probability that the 28th bit received by

your smartphone is the first erroneous bit

1 answer:

alexandr1967 [171]2 years ago

5 0

Probabilities are used to determine the chances of events

The probability that the first erroneous bit is the 28 bit is 0.0076

<h3>How to determine the probability</h3>

The given parameter is:

p =1% --- the probability that a bit has errors

The probability that the first erroneous bit is the 28 bit means that the first 27 bits do not have errors.

So, we have:

$Pr = (1 - p)^{27} * p$

The equation becomes

$Pr = (1 - 1\%)^{27} * 1\%$

Evaluate

$Pr = 0.0076$

Hence, the probability that the first erroneous bit is the 28 bit is 0.0076

Read more about probability at:

brainly.com/question/25870256

You might be interested in

Solve. Good luck! Please do not try to google this.

Elena L [17]

$\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}$
$\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}$
$\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)$
$3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))$
$3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)$
$3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)$
$3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)$
$3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4$
$6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)$
$6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)$

5 0

4 years ago

Evaluate x+1 when x = 2.5

lina2011 [118]

Plug in 2.5 for x and you get

2.5+1 which equals 3.5

5 0

3 years ago

Mark’s salary is $900 per month. What is his yearly salary? $

velikii [3]

His yearly salary is 10,800

6 0

4 years ago

Read 2 more answers

Hhhhhhhhhhhhhhhhhhhhh

musickatia [10]

I cannot see the graph, so I am unsure.

3 0

3 years ago

Read 2 more answers

I have two coins. One is fair, and has a probability of coming up heads of.5. \text {The second is bent, and has a probability o

aniked [119]

Answer:0.625

Step-by-step explanation:

Given

There are two coins one is fair and another is bent

Probability of getting a head on fair coin is 0.5

Probability of getting a head on bent coin is 0.75

If both coins are tossed once

probability of getting at least one of the coin will come up tails is

$P=1-P(both\ heads)$

$P=1-0.5\times 0.75$

$P=1-0.375$

$P=0.625$

6 0

4 years ago