Given - In a science experiment, a mouse is placed in a labyrinth with two different (consider them as A and B) treats hidden in it.
To find - What is the probability that the mouse will find neither of the treats?
Let, P(A) and P(B) are the probabilities for finding first treat and second treat respectively.
So, value of probability that the mouse will find neither of the treats i.e. P(AUB)'
and that we calculate by using below formula.
\(P(AUB) + P(AUB)' = 1\) -------Eq1
1st - The probability that the mouse will find only one of the two treats is 1/5.
P(A∩B') = A only = P(A) - P(A∩B) Similarly P(B∩A') = B only = P(B) - P(A∩B)
P(A∩B') + P(B∩A') = \(\frac{1}{5}\)
P(A∩B') + P(B∩A') = P(A) + P(B) - 2 * P(A∩B) = \(\frac{1}{5}\)
But, P(AUB) = P(A) + P(B) - P(A∩B)
Therefore,
P(A∩B') + P(B∩A')= P(AUB) - P(A∩B) = \(\frac{1}{5}\) -------Eq2
Nothing else we can deduced from this instruction to get the value of probability that the mouse will find neither of the treats i.e. P(AUB)'
Not sufficient.
2nd - The probability that the mouse will find both treats is 3/10.
That is P(A∩B) = \(\frac{3}{10}\) Nothing else so Not sufficient.
Now combining both from equation 2 and P(A∩B) = \(\frac{3}{10}\)
P(AUB) - \(\frac{3}{10}\) = \(\frac{1}{5}\)
P(AUB) = \(\frac{1}{5}\) + \(\frac{3}{10}\)
P(AUB) = \(\frac{1}{2}\)
Now substituting this value in equation 1
P(AUB)' = 1 - \(\frac{1}{2}\)
\(P(AUB)'\) = \(\frac{1}{2}\) That is the probability that the mouse will find neither of the treats.
Answer is C.