Answer:
The required probability is 0.988.
Step-by-step explanation:
Consider the provided information.
Based on a poll, 67% of Internet users are more careful about personal information when using a public Wi-Fi hotspot.
That means the probability of more careful is 0.67
The probability of not careful is: 1-0.67 = 0.33
We have selected four random Internet users. we need to find the probability that at least one is more careful about personal information.
P(At least one careful) = 1 - P(None of them careful)
P(At least one careful) = 1 - (0.33×0.33×0.33×0.33)
P(At least one careful) = 1 - 0.012
P(At least one careful) = 0.988
Hence, the required probability is 0.988.
The result may be higher because of the convenience bias in retrieving the sample. Because the survey subjects volunteered to respond not random.
Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes

.
The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.
The probability of selecting one coin is

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.
P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.
Thus

</span><span>P(B) means that the first envelope contains a quarter AND the
second envelope contains a quarter
</span><span>Thus

Therefore,

Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>

</span><span>

</span><span>
Therefore,

</span>
Aaa^3bxa^2b^3abx^4
=aaa^3a^2abb^3bxx^4
=a^(1+1+3+2+1)b^(1+3+1)x^(1+4)
=a^8^b^5x^5
Answer:
Step-by-step explanation: your friend would get 17$
and you would get 21.25
hope this helpsss ;))))