A bag contains 10 red pens and 6 black pens.
Two pens are chosen from the bag at random without replacement.
Fred claims that the probability that one pen is a red pen and one pen is a black pen can be calculated using \(\left(\dfrac{10}{16}\right)\left(\dfrac6{15}\right)=\dfrac14\).
Explain why Fred’s claim is not true and find the actual probability.[2]
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