In an archery competition, the team consisting of John, David, and Mike will win 1st prize if at least two of them hit the bullseye with their last arrows - Leaving Cert Mathematics - Question 5 - 2016

To find the probability they will win the competition:

1. Calculate Probabilities:
   • Probability of hitting the target:
     • John: $P(J) = \frac{1}{5}$, Miss: $P(J') = 1 - P(J) = \frac{4}{5}$.
     • David: $P(D) = \frac{1}{3}$, Miss: $P(D') = 1 - P(D) = \frac{2}{3}$.
     • Mike: $P(M) = \frac{1}{4}$, Miss: $P(M') = 1 - P(M) = \frac{3}{4}$.
2. Probability of Winning:
   • Winning outcomes:
     • Way 1: $P(Way 1) = P(J)\cdot P(D)\cdot P(M')$
     • Way 2: $P(Way 2) = P(J)\cdot P(D')\cdot P(M)$
     • Way 4: $P(Way 4) = P(J')\cdot P(D)\cdot P(M)$
3. Combine the Probabilities:
   • $P(Win) = P(Way 1) + P(Way 2) + P(Way 4)$
   • Evaluate:
   • $P(Win) = \frac{1}{5} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{5} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{1}{3} \cdot \frac{1}{4}$
   • Simplify to get the final probability.

1. Define Events:
   • $P(A \cap B) = 0.1$
   • $P(B \backslash A) = 0.3$
   • $P(A \backslash B) = x$
2. Using the Union Formula:
   • $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
   • Since $P(B) = 0.3 + x + 0.1$; ($B$ includes $B \backslash A$, $A \cap B$, and $A \backslash B$)
3. Express $P(A)$:
   • $P(A) = P(A \cap B) + P(A \backslash B) = 0.1 + x$
4. Independence Condition:
   • For independence: $P(A \cap B) = P(A)P(B)$
   • Substitute the values: $0.1 = (0.1 + x)(0.4 + x)$
   • Solve for $x$ to find the independent case.