Let:
- P(A) = Probability that the mouse finds the first treat.
- P(B) = Probability that the mouse finds the second treat.
- P(only one treat) = P(A ∩ ¬B) + P(¬A ∩ B)
- P(both treats) = P(A ∩ B)
- P(neither treat) = P(¬A ∩ ¬B)
From the statements:
1. P(only one treat) = 1/5
2. P(both treats) = 3/10
We need to find P(neither treat).
From probability theory:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) + P(¬A ∩ ¬B) = 1
P(¬A ∩ ¬B) = 1 - P(A ∪ B)
Step-by-step solution:
1. Statement (1):
P(only one treat) = P(A ∩ ¬B) + P(¬A ∩ B) = 1/5
2. Statement (2):
P(both treats) = P(A ∩ B) = 3/10
3. Using both statements together:
P(A ∪ B) = P(A ∩ ¬B) + P(¬A ∩ B) + P(A ∩ B) = 1/5 + 3/10
P(A ∪ B) = 2/10 + 3/10 = 5/10 = 1/2
4. Therefore:
P(¬A ∩ ¬B) = 1 - P(A ∪ B) = 1 - 1/2 = 1/2
IMO: C