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3 years ago

Please someone help me!

Mathematics

1 answer:

nekit [7.7K]3 years ago

3 0

Answer:

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Step-by-step explanation:

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I really need help with this guys please help it’s due tomorrow and I’m so frikin dumb

il63 [147K]

Possible 24 impossible 3

5 0

3 years ago

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

Find the length of the side of rhoumbus whose diagonals are of length 6cm and 8cm.

Artist 52 [7]

Answer:

In a rhombus, the diagonals bisect at right angles. That means half the diagonals form a right angle triangle then we can try the Pythagorean theorem. so -

one side of triangle = 6/2 =3 (half of the diagonal)

other side = 8/2 = 4

a^2 + b^2 = c2

3 ^2 + 4^2 = c^2

9+16 = c^2

c^2=25

c = $\sqrt{25}$ = 5