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 at-least one is red.

With the help of answer here, I have computed probability of having both green and this is 4/25.

And with the help of some website having similar question, I computed the probability of having at-least one red as

1 - (4/25) = 21/25.

And this is the right answer (checked from answers given at back of book).

I know that we have to subtract 4/25 but, I am unable to understand that why we have subtracted from 1. Can anyone explain me please ?


Because you have only two options: in the end you have either zero reds, or you have at least one red.

More technically, if we let $R0$ be the event that you don't get any reds, and $R1^+$ the event that you get at least one red, then:

$$R0 + R1^+ = 1$$

because there are no other possible events: how can you not have zero reds, and also not have at least one red?! That's impossible. So, these events cover the whole space of possible outcomes, meaning that they must add up to $1$ or $100$%

Now, in this case to get zero reds is the same as getting two greens, so we know:

$$R0 = \frac{4}{25}$$

From which $R1^+$ readily follows:

$$R1^+ = 1- R0 = \frac{21}{25}$$

  • $\begingroup$ Thank you. I have understand it completely now. $\endgroup$ – Student28 Mar 20 '18 at 20:35
  • $\begingroup$ @Student28 You're welcome! :) $\endgroup$ – Bram28 Mar 20 '18 at 20:53

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