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2 years ago

11

I am terrible with geometry.

1 answer:

Alik [6]2 years ago

7 0

Don't worry I am bad at it too

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HELP ME WITH THIS MATH

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A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a t

USPshnik [31]

Answer:

The probability is $P(K) = \frac{ 28 }{195}$

Step-by-step explanation:

From the question we are told that

The number of green marbles is $n_g = 3$

The number of red marbles is $n_b = 5$

The number of red marbles is $n_r = 7$

Generally the total number of marbles is mathematically represented as

$n_t = n_r + n_g + n_ b$

$n_t = 7 + 3 + 5$

$n_t = 5$

Generally total number of marbles that are not red is

$n_k = n_g + n_ b$

=> $n_k = 3 + 5$

=> $n_k = 8$

The probability of the first ball not being red is mathematically represented as

$P(r') = \frac{n_k}{n_t}$

=> $P(r') = \frac{ 8}{15}$

The probability of the second ball not being red is mathematically represented as

$P(r'') = \frac{n_k - 1}{n_t -1}$

=> $P(r'') = \frac{ 8 -1 }{15-1}$ (the subtraction is because the marbles where selected without replacement )

=> $P(r'') = \frac{ 7 }{14}$

The probability that the first two balls is not red is mathematically represented as

$P(R) = P(r') * P(r'')$

=> $P(R) = \frac{ 8}{15} * \frac{ 7 }{14}$

=> $P(R) = \frac{ 8 }{30}$

The probability of the third ball being red is mathematically represented as

$P(r) = \frac{n_r}{ n_t -2}$ (the subtraction is because the marbles where selected without replacement )

$P(r) = \frac{7}{ 15 -2}$

=> $P(r) = \frac{7}{ 13}$

Generally the probability of the first two marble not being red and the third marble being red is mathematically represented as

$P(K) = P(R) * P(r)$

$P(K) = \frac{ 8 }{30} * \frac{7}{ 13}$

=> $P(K) = \frac{ 28 }{195}$

6 0

2 years ago