A, B, C are events such that \(P(A) = 0.3\), \(P(B) = 0.4\), \(P(C) = 0.8\), \(P(AB) = 0.08\), \(P(AC) = 0.28\) and \(P(ABC) = 0.09\). If \(P(A \cup B \cup C) \geq 0.75\), then show that \(P(BC)\) lies in the interval \(0.23 \leq x \leq 0.48\).
One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 98, 99. If \(x_1\) and \(x_2\) denotes the sum and product of the digits on the tickets, then \(P(x_1 = 9/x_2 = 0)\) is equal to
If two different numbers are taken from the set \(\{0, 1, 2, 3, \ldots, 10\}\), then the probability that their sum as well as absolute difference are both multiples of 4, is
Of the three independent events \(E_1\), \(E_2\), and \(E_3\), the probability that only \(E_1\) occurs is \(\alpha\), only \(E_2\) occurs is \(\beta\), and only \(E_3\) occurs is \(\gamma\). Let the probability \(p\) that none of the events \(E_1\), \(E_2\), or \(E_3\) occurs satisfy the equations \((\alpha - 2\beta)\,p = \alpha\beta\) and \((\beta - 3\gamma)\,p = 2\beta\gamma\). All the given probabilities are assumed to lie in the interval \((0, 1)\). Then \[\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3} = \underline{\hspace{2cm}}.\] (JEE Advanced 2013)
A person goes to office either by car, scooter, bus or train, the probability of which being 1/7, 3/7, 2/7 and 1/7, respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, what is the probability that he traveled by a car?