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Bumek [7]
3 years ago
10

A bag contains 12 counters. 7 of them are white. A counter is taken at random and not replaced. A second counter is taken out of

the bag at random. Calculate the probability that only one of the two counters is white​
Mathematics
1 answer:
grandymaker [24]3 years ago
8 0
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ First calculate how many counters are 'not white'

12 - 7 = 5

We have 5 counters that are 'not white'

The probability of taking a white counter the first time is 7/12

For the next pick, there would be 1 less of the total counters as it is not being replaced

The probability of then taking a 'not white' counter is 5/11

The other way you could still get one white counter is:

The first counter (non white) probability would be 5/12

The second counter (white) probability would be 7/11

Multiply these values:

7/12 x 5/11 = 35/132

5/12 x 7/11 = 35/132

Add these two values together:

35/132 + 35/132 = 70/132

Your answer is 70/132

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

8x + 5y = 18 ------- eqn 1

6x + y = -2 -------- eqn 2

Multiply eqn 2 by -5

-30x -5y = 10 ----- eqn 3

Add eqn 1 and eqn 3 so that y terms gets eliminated

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( + ) --------------

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Divide both sides by -22

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\text{Thus the solution is } x = \frac{-14}{11} \text{ and } y = \frac{62}{11}

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Step-by-step explanation:

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