Two sport shooters shoot : Calculate probabilities Two sport shooters shoot one after the other independently of one another at a target with 10 rings.
Given that all possible outcomes are equally probable, calculate the probability that
(a) both score at least 7 points
(b) at least one scores 9 points or more
(c) the two points differ at most by 1
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I have done the following:
(a) Since the two shoot independently from the other the probability is equal to $\frac{4}{10}\cdot \frac{4}{10}$, since the possible outcome is $7,8,9,10$.
Is that correct?
(b) Do we take the sum of probabilities : propability that S1 scores at least 9 + propability that S2 scores at least 9 ?
(c) How do we calculate that probability?
 A: With the assumption that all outcomes are equally probable as the question states (though it is difficult to see how scoring one point has same probability as scoring ten points), for $(b)$, apply $P(A \cup B) = P(A) + P(B) - P(A\cap B)$ where $A$ and $B$ are events of player $1$ and $2$ scoring at least $9$ points.
For $(c)$, consider player one scoring $1$ point. In that case, player two must score either $1$ point or $2$ points to have at most one point difference with player one. The same is true when player one scores $10$ points. But for score of $2$ points to $9$ points by player one, player two has $3$ choices of score - either one less, equal or one more.
So desired probability $ = 2 \cdot\frac{1}{10} \cdot \frac{2}{10} + 8 \cdot\frac{1}{10} \cdot \frac{3}{10} = \frac{28}{100} = \frac{7}{25}$
Or you can look at all permissible outcomes -
i) Difference of one point in their score - $(1,2), (2,1), (2, 3), (3,2),...(9,10), (10,9)$ which are $9\cdot 2 = 18$ outcomes.
ii) Both have same score - $10$ possible outcomes.
That is $28$ favorable outcomes out of $100$.
