Stats Unit 3

A physics class has 50 students. Of​ these, 16 students are physics majors and 17 students are female. Of the physics​ majors, 6 are female. Find the probability that a randomly selected student is female or a physics major.

(16/50) + (17/50) - (6/50)= 0.54

Question Help A warehouse employs 29 workers on first​ shift, 20 workers on second​ shift, and 15 workers on third shift. Eight workers are chosen at random to be interviewed about the work environment. Find the probability of choosing exactly five first​-shift workers. Use the fundamental counting principle and combinations to find the number of ways of choosing five workers out of the first​-shift workers. A combination is a selection of r objects from a group of n objects without regard to order and is denoted by nCr. The number of combinations of r objects selected from a group of n objects is given by the following formula. nCr=n!(n−r)!r! If eight workers are chosen at random and five are first​-shift ​workers, determine what shifts the remaining workers must be chosen from. The remaining 3 workers can be from either the second shift or third shift. The possible number of ways of choosing 5 first​-shift workers is 29C5. To determine the possible number of ways of choosing 3 workers from either second shift or third ​shift, determine the total number of workers from these two shifts. There are 20+15=35 workers from the second shift or third shift. The possible number of ways of choosing three second shift or third shift workers is 35C3. The fundamental counting principle states that if one event can occur in m ways and a second event can occur in n ​ways, the number of ways the two events can occur in sequence is m•n. ​Thus, the possible number of ways of choosing five first​-shift workers and three second​-shift or third​-shift workers is 29C5•35C3. Find the possible number of ways of choosing five first​-shift workers and three second​-shift or third​-shift workers is 29C5•35C3. ​First, evaluate 29C5. 29C5 = 29!(29−5)!5! = 118,755 Evaluate 35C3. 35C3 = 35!(35−3)!3! = 6545 The calculation for 29C5•35C3 is shown below 29C5•35C3 = 29!(29−5)!5!•35!(35−3)!3! = 118,755•6545 = 777,251,475 To determine the probability of choosing exactly five first​-shift ​workers, divide the combination of choosing five first​-shift workers by the number of combinations of choosing eight workers from any shift. There are 64 workers among all shifts. The calculation for choosing eight workers is shown below. 64C8 = 64!(64−8)!8! = 4,426,165,368 Find the probability of choosing exactly five first​-shift ​workers, rounding to three decimal places. P(five first-shift) = 29C5•35C364C8 = 777,251,4754,426,165,368 = 0.176 ​Therefore, the probability of choosing exactly five first​-shift workers is 0.176.

(29 C 5 * 35 C 3) / 64 C 8

A horse race has 11 entries and one person owns 4 of those horses. Assuming that there are no​ ties, what is the probability that those four horses finish first, second, third, and fourth ​(regardless of​ order)?

11 C 4 11 * 10 * 9 * 8 / 4! = 330 P(E) = 1/330 = .0030

In a certain​ lottery, you must correctly select 6 numbers​ (in any​ order) out of 35 to win. You purchase one lottery ticket. What is the probability that you will​ win?

35 C 6 35 * 34....... * 30 / 6! 1 / ANS

Evaluate the given expression and express the result using the usual format for writing numbers​ (instead of scientific​ notation). 42 P 2 =

42/42 - 2 42/40 42 * 41 = 1722

There are 59 runners in a race. How many ways can the runners finish​ first, second, and​ third?

59 P 3 59 / (59 - 3) 59 / 56 59 * 58 * 57 = 195054

You toss a coin and roll a die. The event​ "tossing tails and rolling a 5 or 2​" is a simple event.

False, the event is not simple because it consists of two possible outcomes

A committee has sixteen members. There are two members that currently serve as the​ board's chairman and ranking member. Each member is equally likely to serve in any of the positions. Two members are randomly selected and assigned to be the new chairman and ranking member. What is the probability of randomly selecting the two members who currently hold the positions of chairman and ranking member and reassigning them to their current​ positions?

Looking for Probability. 16 P 2 16! / 14! = 240 P(E) = 1/240 =

Of the cartons produced by a​ company, 10​% have a​ puncture, 9​% have a smashed​ corner, and 0.6​% have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or a smashed corner.

P(A or B)= P(A) + P(B) - P(A and B) P(Punctured and smashed)= P(Puncture)+P(smashed) - P(both punctured and smashed) P(A or B)= 10% + 9% - 0.6% = 18.4%

In order to conduct an​ experiment, 9 subjects are randomly selected from a group of 49 subjects. How many different groups of 9 subjects are​ possible?

Since the individuals are​ distinct, the order does not​ matter, and an individual cannot be selected more than once for each​ sample, we are counting combinations in this situation. A combination is the selection of r objects from a set of n different objects when the order in which the objects is selected does not matter​ (so AB is the same as​ BA) and an object cannot be selected more than once​ (repetition is not​ allowed). The number of combinations is given by the following formula. nCr=n!r!(n−r!) The value of n is the number of people to choose​ from, so n=49. The value of r is the number of people to​ choose, so r=9. Substitute the numbers into the formula for combinations and simplify. 49C9 =49!9!(49−9)! =49•48•47•46•45•44•43•42•41•40!9!•40! =49•48•47•46•45•44•43•42•419•8•7•6•5•4•3•2•1 =745,520,860,465,920362,880 =​2,054,455,634 The number of different groups which can be obtained is​ 2,054,455,634.

When you calculate the number of permutations of n distinct objects taken r at a​ time, what are you​ counting?

The number of ordered arrangements of n objects taken r at a time. A permutation is an ordered arrangement of objects. The number of different permutations of n distinct objects is n​!.

Determine whether the statement below is true or false. If it is​ false, rewrite it as a true statement. A combination is an ordered arrangement of objects.

The statement is false. A true statement would be​ "A permutation is an ordered arrangement of​ objects."

Determine whether the statement below is true or false. If it is​ false, rewrite it as a true statement. The number of different ordered arrangements of n distinct objects is​ n!.

This statement is true

Determine whether the statement below is true or false. If it is​ false, rewrite it as a true statement. When you divide the number of permutations of 11 objects taken 3 at a time by​ 3!, you will get the number of combinations of 11 objects taken 3 at a time.

This statement is true

Determine whether the statement below is true or false. If it is​ false, rewrite it as a true statement. 7C5 = 7C2

True

A class has 33 students. In how many different ways can five students form a group for an​ activity? (Assume the order of the students is not​ important.)

Use Combination 33 C 5 33 * 32 * 31 * 30 * 29 / 9!

The Probability that A or B will occur is P(A or B)=P(A)+P(B)-P(A and B)

When the two probabilities are added​ together, the probability that they both occur is added​ twice, so it must be subtracted out once to find the correct probability. This is known as the Addition Rule.

When you calculate the number of combinations of r objects taken from a group of n objects what are you​ counting? Give an example.

You are counting the number of ways to select r of the n objects without regard to order. An example of a combination is the number of ways a group of teams can be selected for a tournament.

The first sentence is a statement of the results of several experiments.​ However, the coin is a fair​ coin, so each test is independent and heads and tails occur with equal probability. That means the given statement is false. The correct statement is as shown below.

You toss a fair coin nine times and it lands tails up each time. The probability it will land heads up on the tenth flip is 0.5.

A DJ has a playlist of 19 songs. How many different ways can he arrange the first 5 songs? Since the order of the songs doesn't matter use Permutations to figure out the problem.

n P r n = 19 r = 5 19 P 5 19 / (19 - 5) 19 / 14 19 * 18 * 17 * 16 * 15 = 1395360

Evaluate the given expression and express the result using the usual format for writing numbers​ (instead of scientific​ notation). 40 C 2

nCr = n / (n-r) r 40 C 2 = 40 / (40-2) 2 40 / (38) 2 40 * 39 / 2 * 1 = 780