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3 years ago

11

I really need help!!!

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

$Pr = 6.67\%$

Step-by-step explanation:

Given

$n = 10$ i.e. 10 people

Required

Probability of being next to each other

First, we calculate the total possible arrangements (without any restriction)

$Total = n!$

$Total = 10!$

If the three are to be next to each other,

First, we arrange the three

$r_1 = 3!$

Now, the 3 will be seen as 1; so, we have a total of 8 people i.e. (1 + 7 others)

The arrangement is:

$r_2 =8!$

So, the total arrangement, when they have to be next to one another is:

$r = r_1 * r_2$

$r = 3! * 8!$

The probability is:

$Pr = \frac{r}{n}$

$Pr = \frac{3! * 8!}{10!}$

Expand

$Pr = \frac{3*2*1 * 8!}{10*9*8!}$

$Pr = \frac{3*2*1}{10*9}$

$Pr = \frac{6}{90}$

$Pr = 0.0667$

Express as percentage

$Pr = 0.0667 * 100\%$

$Pr = 6.67\%$

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Help please on this,

n200080 [17]

Answer:

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3 years ago

Read 2 more answers

A total of 2n cards, of which 2 are aces, are to be randomly divided among two players, with each player receiving n cards. Each

klasskru [66]

Answer:

$P(X_s^c|X_F) =0.2$

$P(X_s^c|X_F) =0.31$

$P(X_s^c|X_F) =0.331$

Step-by-step explanation:

From the given information:

Let represent $X_F$ as the first player getting an ace

Let $X_S$ to be the second player getting an ace and

$\sim X_S$ as the second player not getting an ace.

So;

The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;

$P(\sim X_S| X_F) = 1 - P(X_S|X_F)$

$P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}$

Let's determine the probability of getting an ace in the first player

i.e

$P(X_F) = 1 - P(X_F^c)$

$= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}$

$= 1 - \dfrac{n-1}{2(2n-1)}$

$= \dfrac{3n-1}{4n-2} --- (1)$

To determine the probability of the second player getting an ace and the first player getting an ace.

$P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}$ $P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}$

$P(X_sX_F) = \dfrac{n}{2n -1}---(2)$

$P(X_s|X_F) = \dfrac{2}{1}$

$P(X_s|X_F) = \dfrac{2n}{3n -1}$

Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:

$P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}$

$P(X_s^c|X_F) = \dfrac{n-1}{3n -1}$

Therefore;

for n= 2

$P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}$

$P(X_s^c|X_F) = \dfrac{1}{6 -1}$

$P(X_s^c|X_F) = \dfrac{1}{5}$

$P(X_s^c|X_F) =0.2$

for n= 10

$P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}$

$P(X_s^c|X_F) = \dfrac{9}{30 -1}$

$P(X_s^c|X_F) = \dfrac{9}{29}$

$P(X_s^c|X_F) =0.31$

for n = 100

$P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}$

$P(X_s^c|X_F) = \dfrac{99}{300 -1}$

$P(X_s^c|X_F) = \dfrac{99}{299}$

$P(X_s^c|X_F) =0.331$

8 0

3 years ago

Answer both questions please

Anna11 [10]

1.) 37/8

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4 years ago

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agasfer [191]

Answer:

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Step-by-step explanation:

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3 years ago

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Alex17521 [72]

Answer:

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Step-by-step explanation:

√196 = 14 which is a whole no.

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3 years ago