Question

ime, without replacement. Find the probability that the first two marbles selected are not red, and the last marble is red. Round your answer to four decimal places.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 time, without replacement. Find the probability that the first two marbles selected are not red, and the last marble is red. Round your answer to four decimal places.

Answer 1

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

        $P(K) = P(R) * P(r)$

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

=>      $P(K) = \frac{ 28 }{195}$