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a bag contains ten counters four of the counters are red in an experiment three counters are taken from the bag at random and put in a box

calculate the probability that there are exactly two red counters In the box

this is what I have done 4/10*3/10= 3/25 2 - 3/25 = 47/25 is this correct? please explain if not correct

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    The probability can't be greater than $1$! – 2017-01-29
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    Hint: what is the probability that you will draw the exact sequence $RRX$ (where $X$ just means "not red")? What about $RXR$? $XRR$? – 2017-01-29
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    @ lulu, so is the probability of not getting a red is 1- 3/10 = 7/10 if I draw a tree diagram will it help? – 2017-01-29
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    Perhaps, but I think it is more direct to do it the way I sketched. – 2017-01-29
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    Make sure you remember: the probabilities shift after each draw. the probability of getting $R$ first is $\frac 4{10}$. Given that you got an $R$ first, the probability of getting a second $R$ on the second draw is now $\frac 3{9}$ and so on. – 2017-01-29

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Two red counters = $\binom 42$

Non-red counters = $\binom 61$

Probability = $\frac{\binom 42 × \binom 61}{\binom {10}3}$