Mathematics, 11.09.2019 05:30 superbatman9193

Alocal fraternity is conducting a raffle where 50 tickets areto be sold--one per customer. there are three prizes to beawarded. if the four organizers of the raffle each buy oneticket, what is the probability that the four organizers
a) win all of the prizes?
b) win exactly two of the prizes?
c) win exactly one of the prizes?
d) win none of the prizes?
the answers:
a) (4) / (19600)
b) (276) / (19600)
c) (4140) / (19600)
d) (15180) / 19600)

Answers: 2

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Answer from: lyssssasantiago6303

The answers are the same you stated.

The calculations are in the step-by-step explanation

Step-by-step explanation:

There are 3 prizes to be distributed among 50 tickets. The order these prizes are distributed does not matter. So the total number of prizes is a combination of 3 from 50.

The formula for a combination of n from m is:

$C(m,n) = \frac{m!}{n!(m-n)!}$

So, the total number of prizes is:

$T = C(50,3) = \frac{50!}{3!(50-3)!} = \frac{50*49*48*47!}{3!*47!} = 19600$

what is the probability that the four organizers

a) win all of the prizes?

The number of ways that the four organizers can will all of the prizes is a combination of 3 from 4.

$C(4,3) = \frac{4!}{3!1!} = 4$

The probability that the win all of the prizes is the number of ways that they can win all the prizes divided by the total numbers of ways that the prizes can be distributed.

$P = \frac{4}{19600}$

b) win exactly two of the prizes?

The total outcomes(total number of ways that the prizes can be distributed) is 19600.

For them to win exactly two of the prizes, we have a combination of 2 from 4(two organizers win prizes) multiplied by a combination of one from 46(one non-organizers wins a prize), so:

$C(4,2)*C(46,1) = \frac{4!}{2!2!}*\frac{46!}{1! 45!} = 6*46 = 276$

The probability that they win exactly two of the prizes is

$P = \frac{276}{19600}$

c) win exactly one of the prizes?

The total outcomes(total number of ways that the prizes can be distributed) is 19600.

For them to win exactly one of the prizes, we have a combination of 1 from 4(one organizer wins a prize) multiplied by a combination of two from 46(two non-organizers win prizes), so:

$C(4,1)*C(46,2) = \frac{4!}{1!3!}*\frac{46!}{2! 44!} = 4*1035 = 4150$

The probability that they win exactly one prize is

$P = \frac{4150}{19600}$

d) win none of the prizes?

The total outcomes(total number of ways that the prizes can be distributed) is 19600.

For them to win none of the prizes, we have a combination of 3 from 46(3 prizes distributed among 46 non-organizers). So:

$C(46,3) = \frac{46!}{43!3!} = 15180$

The probability that they don't win any prize is:

$P = \frac{15180}{19600}$

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

Answer from: Quest

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

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