Mathematics, 26.02.2020 17:00 kimezzell18

A bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken out of the bag at random. calculate the probability that only one of the counters is white. give your answer as a fraction

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Answer from: mem8163

We want to find the probability that, after taking 2 counters with no replacement, that only one of these is white. We will see that the probability is 6/11.

So we know that there are 11 counters, 5 of them are white, 6 are not white.

Let's assume that first we take the white one, the probability is equal to the quotient between the number of white counters and the total number of counters, so we have:

p = 5/11

Now for the second one, we don't want to take a white one, so here the probability is equal to the quotient between the number of non-white counters (6) and the total number of counters (now there are 10, because we took one)

q = 6/10

The joint probability is the product of the individual probabilities, this is:

P = p*q

And, also notice that there are 2 permutations:

first white, then non-white.first non-white, then white.

So we need to add a factor 2.

P = 2*p*q = 2*(5/11)*(6/10) = 6/11

The probability is 6/11

If you want to learn more, you can read:

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Answer from: aarionna8

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

Probabilities

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

Answer from: Quest

show us the graphs dude

step-by-step explanation:

Answer from: Quest

answer: ur v dumb

step-by-step explanation:

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