Consider Again Problem 2 1 3 Where Two Cards Are Drawn
MAT 312: Probability and Statistics for Middle School Teachers
Spring 1999
9:35 - 10:fifty am TR STV 350A
Dr. Roger Day (day@ilstu.edu)
Consider the experiment of selecting a carte from an ordinary deck of 52 playing cards. Determine the probability of each outcome.
1.
A confront carte is drawn.
12/52 = 3/13
2.
A red card or a card showing a 5 is drawn.
P(Scarlet OR five) = P(Crimson) + P(5) - P(Blood-red AND 5) = 26/52 + four/52 - 2/52 = 28/52 = 7/13
3.
A non-face up card or a seven is drawn.
P(non-face OR 7) = P(not-face) + P(7) - P(non-confront AND 7) = 40/52 + 4/52 - 4/52 = 40/52 = 10/13
4.
A card drawn is neither a king nor a spade.
There are 4 kings and there are 13 spades, with one that is both a king and a spade, so there are 4 + 13 - 1 = sixteen cards that are either kings or spades. That leaves 52-xvi=36 cards that are neither a king nor a spade. Our desired probability is 36/52 = 9/thirteen.
5.
A card that is a black face up card is drawn.
There are 6 black face cards, so the probability is half-dozen/52 = iii/26.
6.
A carte du jour that is not a face card is fatigued.
We know from question (1) that P(face carte du jour) = iii/thirteen. Because the event "non-face card" is the complement to "face card," P(non-face card) = i - P(face up menu) = 1 - 3/13 = ten/13.
A box contains iii ruby-red balls and ii white balls. A brawl is selected at random from the box, its colour is recorded, and the ball is replaced. A second brawl is and so selected at random and its color is recorded. The outcome associated with this type of selection is an ordered pair (first drawn 2nd draw). An instance result is (ruby-red,white).
seven.
List the sample space for this experiment.
At that place are four ordered pairs in the sample space, where R stands for a red brawl drawn and W represents a white ball fatigued: {(R,R),(R,Due west),(W,R),(West,W)}.
viii.
Determine the probability that both balls are red.
Because we are drawing with replacement, the outcome of the 2nd draw does not depend on the outcome of the first. Therefore, the events are independent and we tin can mutliply to determine the desired probability: P(R,R) = P(R)*P(R) = iii/5 * 3/5 = ix/25.
9.
Determine the probability that both assurance are white.
Using the aforementioned justification equally in question (8), P(W,W) = P(W)*P(Westward) = 2/5 * ii/5 = 4/25.
x.
Explicate why the probabilities adamant for (eight) and (9) exercise not sum to i.
The events in questions (viii) and (9) do non represent all possible outcomes of the experiment, for we could get (R,W) or (W,R) every bit well.
| The table of information was recently nerveless at the Jamestown Schoolhouse dining hall. Use this information to calculate the probabilities indicated for someone picked at random from the school's student population. For the questions here, probabilities are adamant by creating a ratio of "favorable cases" to "total cases." |
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11.
P(male)
386/791
12.
P(10th grader)
200/791
13.
P(male | 9th grader)
91/192
14.
P(9th grader | male)
91/386
xv.
P(junior or senior)
399/791
16.
P(11th grader and female)
98/791
17.
Suppose we know that for event A, P(A) = 0.65. If ~A represents the event complementary to A, what is P(~A)?
P(~A) = 1 - P(A) = 1 - 0.65 = 0.35
eighteen.
An experiment has three possible outcomes, 10, Y, and Z. If P(Z) = P(X) and P(Y) = three·P(X), make up one's mind P(10), P(Y), and P(Z).
Let P(X) = a. Then P(Z) = a and P(Y) = 3a. We must have P(Ten) + P(Y) + P(Z) = ane, because X, Y, and Z comprise the unabridged sample space. Then P(Ten) + P(Y) + P(Z) = 1 implies that a + 3a + a = one. Solving for a yields a = one/5. This gives u.s. P(Ten) = P(Z) = 1/v and P(Y) = 3/5.
A family unit has three children, ages 5, vii, and 10. If nosotros assume that the probability of giving nascency to a boy is the same every bit the probability of giving nascence to a girl, decide the following probabilities.
For these problems, nosotros use the post-obit information, where B represents a boy and G represents a daughter. The prepare of all outcomes is a set up of 8 equally likely outcomes, shown in nascence order from youngest to oldest:
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19.
P(three girls)
one/8
20.
P(at least two boys)
4/8 = ane/2
21.
P(the oldest is a daughter)
4/viii = one/2
22.
P(two children are of the same gender)
6/8 = iii/4 (interpreted as "exactly two")
23.
P(there are an equal number of boys and girls)
0: This cannot happen in a family unit of three.
Three different gaming machines were on showroom at a local casino. A sign on each machine showed its sample space (in dollars) and the probability of each guaranteed output in the sample infinite.
24.
If yous wanted to select the car that produced the largest average (mean) output over a long time menstruum, which machine would you select? Explain.
Select Machine A. Over the long haul, here's what each car volition practise, expressed as a weighted average using the sum of the product of each issue with its probability. This is called expected value:
- Car A: (2/seven)*one + (iii/vii)*3 + (2/7)*5 = 21/seven = iii
- Machine B: (v/ix)*0 + (4/9)*6 = 24/9 = 8/3
- Car C: (one)*ii = two
25.
Suppose that you and ii friends are beginning to play the three machines. If you have first choice on which automobile you'll play, which of the three would yous choose? Explain.
I translate this to mean I want to cull the machine whose most probable outcome is greatest. The well-nigh likely event of Car A is 3, of B is six, and of C is 2. I'd choose Machine B.
26.
A fair dice is rolled three consecutive times. What is the probability that the digit sum of the 3 rolls is fifteen or larger?
There are 6*half-dozen*half-dozen = 216 possible outcomes when a fair die is rolled three times. Nosotros at present demand to determine how many of those outcome in a sum of 15 or more.
The possible sums are 18, 17, 16, and 15. Nosotros list beneath the means to become each of those. Each entry is an ordered triple showing the result on the first, second, and 3rd roll, respectively. For example, (half dozen,5,vi) represents getting a vi on the starting time roll, a five on the second, and a 6 on the third.
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| (six,iii,6) (iii,vi,6) (half-dozen,v,4) (six,4,five) (five,6,4) (5,4,half-dozen) (four,6,5) (4,5,6) (five,v,v) | (6,4,six) (4,6,half dozen) (6,5,5) (v,6,5) (5,5,6) | (vi,5,6) (5,vi,six) | |
This list shows 20 ways to get the desired sums. The probability we seek is twenty/216 = v/54.
27.
A fair die is rolled six times. Determine the probability that each of the six equally probable outcomes appears exactly in one case in those six rolls.
Again nosotros rely on the archetype probability ratio, comparing favorable cases to the total number of cases.
At that place are six*vi*6*half-dozen*half dozen*6=6^6 (6 raised to the sixth power) different outcomes when a fair die is rolled 6 times. The favorable outcomes are just the means to arrange the digits {1,2,three,4,5,six}, considering each must appear one time. This can exist done in P(6,half dozen) = 6! means.
Therefore, the desired probability is 6!/(6^6) = 5/324 or approximately 0.0154.
A production line is equipped with two quality-control bank check points that tests all items on the line. At bank check point #i, 10% of all items failed the test. At check point #2, 12% of all items failed the test. Nosotros also know that 3% of all items failed both tests.
28.
If an item failed at check point #ane, what is the probability that it as well failed at check bespeak #2?
In symbolic form, we are asking for P(neglect #2 | fail #1). Using the formula for determining provisional probability, P(fail #2 | neglect #1) = P(fail #2 AND fail #1)/P(fail #1) = (0.03)/(0.10) = 3/x.
29.
If an detail failed at cheque point #2, what is the probability that it likewise failed at bank check indicate #1?
Hither nosotros are asking for P(neglect #1 | neglect #2). Using the formula, P(neglect #1 | fail #two) = P(fail #ane AND fail #2)/P(fail #2) = (0.03)/(0.12) = 3/12 = ane/four.
30.
What is the probability that an item failed at check indicate #1 or at check signal #2?
P(neglect #1 OR fail #2) = P(fail #1) + P(neglect #2) - P(fail #i AND neglect #2) = 0.10 + 0.12 - 0.03 = 0.19.
31.
What is the probability that an item failed at neither of the check points?
We are asking for P(no failure). The event "no failure" is the complement of the result "failure in at least i exam." Nosotros know from question (30) that P(failure in at least one examination) = 0.nineteen, and so P(no failure) = ane - 0.19 = 0.81.
George knows that a rare disease, D, within his family can be passed on to his children, and that the probability is 0.ten that the inherited affliction will be passed on to a kid. We signify this at P(D) = 0.10.
32.
Make up one's mind the probability that none of George's three children inherit the disease from George.
Utilize ~D to correspond the complement of event D. Then ~D represents "did not pass on the affliction" and because P(D) = 0.10, P(~D) = 0.90.
We want P(~D kid #i AND ~D child #2 AND ~D kid #3). We assume that these are contained events, that is, that whether the disease has passed to one child doesn't modify the probability of it being passed to some other. Bold this, nosotros accept
P(~D child #1 AND ~D child #two AND ~D child #3) = P(~D child #ane)*P(~D child #2)*P(~D child #iii) = (0.9)*(0.9)*(0.9) = 0.729, the desired probability.
33.
If P(D) =x, make up one's mind the largest value of x possible so that the solution to (32) would be greater than or equal to 90%.
Using the information, symbolism, and strategy from question (32), we seek ten so that x*x*x is greater than or equal to 0.9. By guess-and-check or by using the cube-root function on a calculator, nosotros determine that x = 0.9655, rounded to the nearest ten-thousandth.
Suppose the Atlanta Braves and the New York Yankees run into in the next Major League Baseball Earth Series, where the two teams play until ane team has won four games. Let B represent the result that the Braves win a game and let Y represent the event that the Yankees win a game, with P(B) = 0.4 and P(Y) = 0.6). Assume that these probabilities do not modify throughout the series. Determine the following probabilities for this World Serial.
For the questions beneath, nosotros use 2 strategies: (1) Employ multiplication of consecutive probabilities to make up one's mind the probability of a specific sequence of games. (2) Count in some way the number of means a certain event can occur. Question (34) illustrates employ of strategy #1 and question (35) uses both strategies.
We use the events B and Y above, writing them in sequence to evidence a string of results. for instance, BYBYY represents the Braves winning game 1 and game 3 and the Yankees winning games two, iv, and 5.
Nosotros besides interpret the statement in a higher place, "Assume that these probabilities do not modify throughout the series," to indicate that the outcome of one game does not influence the outcome of another.
34.
P(Atlanta wins no games)
This will only happen if the Yankees win the first 4 games. Thus we seek P(YYYY) and past the independence of game results, P(YYYY)=P(Y)*P(Y)*P(Y)*P(Y)=0.6^4 = 0.1296.
35.
P(serial is tied 2 games each after the outset 4 games)
In that location are half dozen ways this tin happen, each with the same probability:
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The probability of each is (0.6)^ii*(0.4)^2 = 0.0576. Multiply this by half-dozen to become the desired probability: P(series is tied two games each after the first 4 games) = 0.3456
36.
P(the serial ends in 5 games)
To end in 5 games, the serial record must be four wins and 1 loss for i of the teams. The probability of each will change depending on which team has 4 wins:
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| = 4(0.four)^4(0.vi) = 0.06144 | |||
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| = iv(0.6)^4(0.four) = 0.20736 | |||
Because the two events (Braves win 4-1, Yankees win 4-1) are mutually sectional, nosotros add the probabilities of each event. Therefore, P(the series ends in v games) = 0.06144 + 0.20736 = 0.2688.
37.
P(if the series lasts seven games, Atlanta will win).
We seek P(Atlanta wins | serial last 7 games). This is a provisional probability statement and we use the conditional formula here.
P(Atlanta wins | series last seven games) = P(Atlanta wins AND series last 7 games)/P(series lasts seven games)
In that location are forty dissimilar win-lose sequences that lead to a seven-game series. In 20 of these sequences, Atlanta wins 4 games to iii and in the other 20, the Yankees win 4 games to 3.
Each of the 20 ways that Atlanta wins the series in seven games has the same probability: (0.4)^4(0.6)^3 = 0.0055296. So P(Atlanta wins AND serial concluding 7 games) = 20*(0.4)^4(0.6)^3 = 20*0.0055296 = 0.110592.
Each of the 20 means that the Yankees win the series in 7 games has the same probability: (0.six)^four(0.4)^3 = 0.0082944. And then P(Yankees win AND series last vii games) = 20*(0.6)^4(0.four)^3 = xx*0.0082944 = 0.165888.
Because the event "Yankees win in 7" and the event "Braves win in 7" are mutually sectional, we add the two results to determine P(series terminal 7 games). This is 0.165888 + 0.110592 = 0.27648 = P(series last 7 games). This says, by the manner, that nether these conditions and assumptions, about 25% of the time the series will last vii games.
We can at present compute the desired conditional probability:
P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last vii games)/P(series lasts 7 games) = (0.110592)/(0.165888 + 0.110592) = 0.iv. Does this seem similar a surprising result? Given that we get to game 7, nosotros can call up of the series humid downwards to one game, and the probability that atlanta wins that one game is merely P(B) = 0.4.
Another way to think about (0.110592)/(0.165888 + 0.110592) is top call up of it as a weighted outcome: P(effect A and status #1) compared to the sum of the probabilities of all possible results nether status #1. Hither, there were just 2 results possible under the status that we get to game 7: Either the Braves win or the Yankees win.
Other situations may extend that. Suppose nether a sure condition, 10 dissimilar things - - all mutually exclusive - - could happen. Then the probability that the outset of those unlike things does happen under the given condition is but the probability that that first thing happens and the condition happens compared to the sum of the ten probabilities for the 10 unlike things that could happen with the condition in place.
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Source: https://math.illinoisstate.edu/day/courses/old/312/assign/probproblemssoln.html
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