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ycow [4]
3 years ago
11

A bag contains 10 marbles,7 of which are black and 3 are red.three marbles are drawn at one after the other.find the probability

of choosing at most two black marbles.​
Mathematics
1 answer:
S_A_V [24]3 years ago
8 0
<h3>Answer:  17/24</h3>

==========================================================

Explanation:

We have these four cases or possible outcomes

  • Case 1) We select 0 black marbles and 3 red marbles.
  • Case 2) We select 1 black marble and 2 red marbles.
  • Case 3) We select 2 black marbles and 1 red marble.
  • Case 4) We select 3 black marbles and 0 red marbles.

Let's calculate the probability for case 4.

There are 7 black marbles out of 10 total. The probability of picking black is 7/10. If no replacement is made, then 6/9 is the probability of picking black again (subtract 1 from the numerator and denominator separately). Finally, 5/8 is the probability of getting black a third time.

The probability of getting 3 black marbles in a row is

(7/10)*(6/9)*(5/8) = (7*6*5)/(10*9*8) = 210/720 = 7/24.

That fraction 7/24 means that if you had 24 chances, then you expect about 7 of them will lead to getting three black marbles in a row (aka case 4). Therefore, 24-7 = 17 occurrences are expected where we get cases 1 through 3 occur in some fashion (pick one case only).

Notice how cases 1 through 3 encapsulate the phrasing "at most 2 black marbles" which is another way of saying "2 black marbles is the highest we can go".

So that's why the answer is 17/24.

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Fiesta28 [93]

Answer:

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On putting values of P\,,\,T\,,\,S.I in formula , we get S.I. = \frac{P\times R\times T}{100}

56.24 = \frac{1600\times R\times 1}{600}\\R=\frac{56.24\times 600 }{1600}=\frac{703\times 3}{100}=\$21.09

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7 0
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