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Alika [10]
3 years ago
9

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

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.
Mathematics
1 answer:
USPshnik [31]3 years ago
6 0

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}

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A closet contains n pairs of shoes. If 2r shoes are chosen at random, (where 2r < n), what is the probability that there will
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We are choosing 2
2
r
shoes. How many ways are there to avoid a pair? The pairs represented in our sample can be chosen in (2)
(
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ways. From each chosen pair, we can choose the left shoe or the right shoe. There are 22
2
2
r
ways to do this. So of the (22)
(
2
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equally likely ways to choose 2
2
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shoes, (2)22
(
n
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r
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2
2
r
are "favourable."

Another way: A perhaps more natural way to attack the problem is to imagine choosing the shoes one at a time. The probability that the second shoe chosen does not match the first is 2−22−1
2
n
−
2
2
n
−
1
. Given that this has happened, the probability the next shoe does not match either of the first two is 2−42−2
2
n
−
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. Given that there is no match so far, the probability the next shoe does not match any of the first three is 2−62−3
2
n
−
6
2
n
−
3
. Continue. We get a product, which looks a little nicer if we start it with the term 22
2
n
2
n
. So an answer is
22⋅2−22−1⋅2−42−2⋅2−62−3⋯2−4+22−2+1.
2
n
2
n
⋅
2
n
−
2
2
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2
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4
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This can be expressed more compactly in various ways.
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