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A bag contains ten counters of which six are red and four are green. A counter is chosen at random; its colour is noted and it is replaced in the bag. A second is then chosen at random. Find the probability that both counters are red.

How we would start this question ? What would be the sample space ? I have completed the questions with dice and card deck but, this question is different from those questions. Can anyone help me please ?

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  • $\begingroup$ What's the probability that the first counter selected is red? $\endgroup$ – quasi Mar 18 '18 at 11:42
  • $\begingroup$ @quasi that would be 6/10 which is 3/5. $\endgroup$ – Student28 Mar 18 '18 at 11:45
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    $\begingroup$ So consider the events $\text{RR},\,\text{RG},\,\text{GR},\,\text{GG}$. Can you compute probabilities for each of those $4$ events? $\endgroup$ – quasi Mar 18 '18 at 11:46
  • $\begingroup$ @quasi Yes. Thank you. I am going to complete it now. $\endgroup$ – Student28 Mar 18 '18 at 11:51
  • $\begingroup$ You only care about the event $\text{RR}$. I was just suggesting to compute all $4$ probabilities to help understand the experiment (and the associated sample space). $\endgroup$ – quasi Mar 18 '18 at 11:56
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To make everybody understand, I will not only consider the case of the problem, but also for the case "Counter is not put back in the bag".

If the counter is put back in the bag, then the probability of picking a red counter on each time is independant from each other, so the probability of having picked a red counter two times is: $\frac{6}{10}\times \frac{6}{10}=36$%.

If the counter is not put back in the bag, then the probability of picking a red counter on each time is dependant, in this case:

  • The probability of having picked a red counter on the first pick is $6$ (red counters) out of $10$ (number of counters in total).

  • The probability of having picked a red counter on the second pick is $5$ (red counters remaining) out of $9$ (number of counters in total remaining).

If the problem is changed like that, the probability of having picked both red counters would be $\frac{6}{10} \times \frac{5}{9}=33.33$%.

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