Seems illogical probability question Two basketball players each throw one shot to the basket. One of the players probability that he will score is 0.7. And both players getting ball in the basket probability is 0.82. P(A) = 0.7 P(AB) = 0.82 And need to find out P(B). Others guys probability. But what I cant seem to understand is that both players scoring probability is higher than A guys probability. What I tried to do was: P(AB) = P(A) * P(B) 0.82 = 0.7 * P(B) P(B) = 0.82/0.7 => larger than 1 so is impossible. Explain how that is possible and how can I calculate it? Answer is 0.4

Based on the answer, the question should say that the probability of _at least one_ player getting the ball in the basket is 0.82.

Edit: skv has suggested that I write out the calculation:

$$\mathbb{P}(A)=0.7$$ $$\mathbb{P}(A \cup B)=0.82$$ $$\mathbb{P}(A \cup B)=1-(1-\mathbb{P}(A))(1-\mathbb{P}(B))$$ $$0.82=1-0.3(1-\mathbb{P}(B))$$ $$0.3(1-\mathbb{P}(B))=0.18$$ $$1-\mathbb{P}(B)=0.6$$ $$\mathbb{P}(B)=0.4$$

Dilip's digit reversal suggestion would also give an answer of 0.4.