Consider the experiment of tossing a fair coin 3 times. For each coin, the possible outcomes are heads or tails.(a) List the equ
1 answer:
Answer:
a)
H - H - H
H - H - T
H - T - H
H - T - T
T - H - H
T - H - T
T - T - H
T - T - T
b)
0.125 = 12.5% probability that all three coins come up heads.
0.875 = 87.5% probability that there will be at least one tail.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Question a:
Considering H for heads, T for tails:
H - H - H
H - H - T
H - T - H
H - T - T
T - H - H
T - H - T
T - T - H
T - T - T
(b) What is the probability that all three coins come up heads?
One outcome(H - H - H) out of 8, so:
0.125 = 12.5% probability that all three coins come up heads.
Notice that the complement of the event "3 heads" is "at least one tail." Use this information to compute the probability that there will be at least one tail.
Sum of these probabilities is 100%, so:
1 - 0.125 = 0.875
0.875 = 87.5% probability that there will be at least one tail.
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Answer:
Is a very improbable event.
Step-by-step explanation:
We want to calculate the probability that the total weight exceeds the limit when the average weight x exceeds 6000/100=60.
If we analyze the situation we this:
If represent the 100 random beggage weights for the n=100 passengers . We assume that for each
for each
the distribution assumed is normal with the following parameters
.
Another important assumption is that the each one of the random variables are independent.
1) First way to solve the problem
The random variable S who represent the sum of the 100 weight is given by:
The mean for this random variable is given by:
And the variance is given by:
And the deviation:
So we have this distribution for S
On this case we are working with the total so we can find the probability on this way:
2) Second way to solve the problem
We know that the sample mean have the following distribution:
If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:
And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.