When messaging a message in Morse code, the chances of a sign being received well are 95%. What is the probability that 50 words of 6 characters will not receive more than 10 wrong words?
can someone answer pls or just say if u don't know..... then i dont have to wait
well the recieved well is not true rather depends on your memory, the probability is about 1 in 37 each mess up even just a mess up on a letter.
To find the probability that 50 words of 6 characters will not receive more than 10 wrong words, we can use the binomial distribution formula.
The formula for the probability of k successes in n trials, where the probability of success is p, is:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
In this case, a "success" would be a word received well (95% accuracy). So the probability of success (p) is 0.95, and the probability of failure (1-p) is 0.05.
We want to find the probability that there are 10 or fewer wrong words, so we need to calculate the sum of probabilities for k = 0 to 10:
P(0) + P(1) + ... + P(10)
Let's calculate this using the formula:
P(0) = C(50, 0) * 0.95^0 * 0.05^50
P(1) = C(50, 1) * 0.95^1 * 0.05^49
P(2) = C(50, 2) * 0.95^2 * 0.05^48
...
P(10) = C(50, 10) * 0.95^10 * 0.05^40
To calculate these probabilities, we need to use the combination formula C(n, k):
C(n, k) = n! / (k! * (n-k)!)
Calculating all these probabilities and summing them will give us the final probability.
Note: This calculation can be complex. It can be simplified by using a cumulative distribution function (CDF) for the binomial distribution, which sums up the probabilities for k = 0 to 10 automatically. However, since you are asking for an explanation of the calculations, I have shown the step-by-step approach.