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Amanda [17]
3 years ago
10

According to Internet security experts, approximately 90% of all e-mail messages are spam (unsolicited commercial e-mail), while

the remaining 10% are legitimate. A system administrator wishes to see if the same percentages hold true for the e-mail traffic on her servers. She randomly selects e-mail messages and checks to see whether or not each one is legitimate. (Unless otherwise specified, round all probabilities below to four decimal places; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.) Assuming that 90% of the messages on these servers are also spam, compute the probability that the first legitimate e-mail she finds is the fifth message she checks:
Mathematics
1 answer:
KonstantinChe [14]3 years ago
3 0

Answer:

0.0656

Step-by-step explanation:

For each message, we have these following probabilities:

90% probability it is spam.

10% probability it is legitimate.

Compute the probability that the first legitimate e-mail she finds is the fifth message she checks:

The first four all spam, each with a 90% probability.

The fifth legitimate, with a 10% probability.

P = (0.9)^{4} \times 0.1 = 0.0656

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Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on
Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

  • If he is given with two kings.
  • If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by =  \frac{^{3}C_2 }{^{47}C_2}   {∵ one king is already there}

              =  \frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}           {∵ ^{n}C_r = \frac{n!}{r! \times (n-r)!} }

              =  \frac{3}{1081}  =  0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by =  \frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}

                                                                   =  \frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}

                                                                   =  \frac{6}{1081}  =  0.0055

Now, probability that he ends up with a full house = \frac{3}{1081} + \frac{6}{1081}

                                                                                    =  \frac{9}{1081} = <u>0.0083</u>.

3 0
3 years ago
Read 2 more answers
Suppose you earned 8t-3 on monday and 6t+5 dollars on tuesday. what were your total earnings? simplify your answer
Nadusha1986 [10]
Let
A=8t-3
B=6t+5

we know that
<span>total earnings=A+B
</span>so
total earnings=(8t-3)+(6t+5)------> (8t+6t)+(-3+5)------> (14t+2) <span>dollars

the answer is
</span>(14t+2) dollars<span>

</span>
6 0
3 years ago
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