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Name: _______________________________________ Date: _____________ Week 3 – Homew

ID: 3289261 • Letter: N

Question

Name: _______________________________________                  Date: _____________

Week 3 – Homework

1. Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = 1).

a. Describe the outcomes of this experiment. List the elements of the sample space.

b. What is the probability of obtaining a total of 5 for the two cards?

c. Let Abe the event “total card value is 5 or less.” Find P(A) and P(Ac).

2. Three coins are dropped on a table.

a. List all possible outcomes in the sample space.

b. Find the probability associated with each outcome.

c. Let Abe the event “exactly 2 heads.” Find P(A).

d. Let Bbe the event “at most 1 head.” Find P(B).

e. Let Cbe the event “at least 2 heads.” Find P(C).

f. Are the events Aand Bmutually exclusive? Find P(Aor B).

g. Are the events Aand Cmutually exclusive? Find P(Aor C).

3. A market research company surveyed consumers to determine their ranked preferences of energy drinks among the brands Monster, Red Bull, and Rockstar.

a. What are the outcomes of this experiment for one respondent?

b. What is the probability that one respondent will rank Red Bull first?

c. What is the probability that two respondents will both rank Red Bull first?

4. Roulette is played at a table similar to the one in the figure shown below. A wheel with the numbers 1 through 36 (evenly distributed with the colors red and black) and two green numbers 0 and 00 rotates in a shallow bowl with a curved wall. A small ball is spun on the inside of the wall and drops into a pocket corresponding to one of the numbers. Players may make 11 different types of bets by placing chips on different areas of the table. These include bets on a single number, two adjacent numbers, a row of three numbers, a block of four numbers, two adjacent rows of six numbers, and the five number combinations of 0, 00, 1, 2, and 3; bets on the numbers 1–18 or 19–36; the first, second, or third group of 12 numbers; a column of 12 numbers; even or odd; and red or black. Payoffs differ by bet. For instance, a single?number bet pays 35 to 1 if it wins; a three number bet pays 11 to 1; a column bet pays 2 to 1; and a color bet pays even money. Define the following events: C1 = column 1 number, C2 = column 2 number, C3 = column 3 number, O = odd number, E = even number, G= green number, F12 = first 12 numbers, S12 = second 12 numbers, and T12 = third 12 numbers.

a. Find the probability of each of these events.

b. Find P(Gor O),P(Oor F12), P(C1 or C3), P(Eand F12), P(Eor F12),P(S12 and T12), and P(Oor C2).

03-07

5. An airline tracks data on its flight arrivals. Over the past six months, 25 flights on one route arrived early, 150 arrived on time, 45 were late, and 30 were cancelled.

a. What is the probability that a flight is early? On time? Late? Cancelled?

b. Are these outcomes mutually exclusive?

c. What is the probability that a flight is either early or on time?

03-07

Explanation / Answer

1. As there are 4 pairs of each number from ace to 10 the possibility will be all from 2( by adding 2 ace) to 20 ( adding two 10)

5 can be obtained by adding 4and 1 or 3&2

Probability of getting 4 in the first case is 4/40 and getting 1 in the other card is 4/39 as 1 card is removed. So the probability of getting one 4 and one ace is (4*4)/40*39=4/390

Similarly again in second case probability comes out to be 4/390

So total is 4/390 + 4/390 = 8/390 =4/195

C.For third part of the question

Value should be 5 or less

Minimum value can be 2 which can be obtained by adding two aces probability is 4/40 * 3/39= 3/390

3 can be obtained by adding one ace and one two

Probability for which is the same as above 4/390

4 can be obtained by adding two 2s or one 3 and one ace

Probability of getting two 2s is 3/390 and probability of one 3 and one ace is 4/390 so total probability is 3/390+4/390 =

7/390

Probability of getting 5 is already calculated to be 8/390

So total probability of getting 5 or less is

3/390+4/390+ 7/390+8/390= 22/390= 11/195

Please(Ac) is 1-11/195= 84/195

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