Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. The probability of A's winning if A starts first is

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is four. The probability that it is actually four is

A can solve 90% problems and B can solve 70% problems of the book. A problem is selected at random from the book. The probability that the problem is solved, is equal to

A random variable X has the following probability distribution:

The variance of the X will be:

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first :

A problem is given to 3 students A,B and C, whose probabilities of solving are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively, then

(A). P(Problem solved) = $\frac{3}{4}$

(B). P(Problem not solved) = $\frac{1}{4}$

(C). P(Problem solved by exactly one student) = $\frac{7}{24}$

(D). P(Problem solved by at least 2 students) = $\frac{17}{24}$

Choose the correct answer from the options given below:

Let X denotes the number of heads in a simultaneous toss of three coins, then $P(0 < X < 3)$ is