For the following distribution, the mean deviation about the median is
$$ \begin{array}{cccccccc} \hline x_i & 6 & 12 & 18 & 24 & 30 & 36 & 42 \\ \hline f_i & 4 & 7 & 9 & 18 & 15 & 10 & 5 \\ \hline \end{array} $$
If a man throws a die until he gets a number bigger than 3 , then the probability that he gets a 5 in his last throw is
A diagnostic test has the probability 0.95 of giving a positive result when applied to a person suffering from a certain disease and a probability 0.10 of giving a positive result when given to a non-sufferer. It is estimated that $0.5 \%$ of the population are suffering from the disease. If this test is now administered to a person from this population about whom there is no information relating to the incidence of this disease and the test gives a positive result, then the probability that he is a sufferer, is
Consider the following statements
Assertion (A) If $P_1, P_2, P_3$ are probability of happening of three independent events, then probability of happening of atleast one of them is $1-\left[\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)\right]$
Reason (R) For any three independent events $A, B$ and $C$
$$ \begin{array}{r} P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A) P(B)-P(A) P(C) -P(B) P(C)+P(A) P(B) P(C) \end{array} $$
The correct option among the following is
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