All the jacks, queens, kings, and aces of a regular 52 cards deck are taken out. The 16 cards are thoroughly shuffled and my opponent, a person who always tells the truth, simultaneously draws two cards at random and says, "I hold at least one ace". The probability that he holds two aces is
Three ships \(A\), \(B\), and \(C\) sail from England to India. If the ratio of their arriving safely are 2:5, 3:7, and 6:11, respectively, then the probability of all the ships for arriving safely is
Two players $P_1$ and $P_2$ play a game. Each player rolls a die once. If $x>y$: $P_1$ scores 5, $P_2$ scores 0. If $x=y$: each scores 2. If $x<y$: $P_1$ scores 0, $P_2$ scores 5. Match: I) $P(X_2\ge Y_2)$; II) $P(X_2>Y_2)$; III) $P(X_3=Y_3)$; IV) $P(X_3>Y_3)$ with P) 3/8, Q) 11/16, R) 5/16, S) 355/864, T) 77/432.
An event \(X\) can take place in conjuction with any one of the mutually exclusive and exhaustive events \(A\), \(B\) and \(C\). If \(A\), \(B\), \(C\) are equiprobable and the probability of \(X\) is 5/12, and the probability of \(X\) taking place when \(A\) has happened is 3/8, while it is 1/4 when \(B\) has taken place, then the probability of \(X\) taking place in conjuction with \(C\) is
Question nos. 646 to 648Let \(X = \{1, 2, 3, \ldots, 10\}\). \(A\), \(B\), \(C\) are three sets such that \(A \subseteq X\), \(B \subseteq X\) and \(C \subseteq X\).Column-1: Contains types of three subsets of \(X\).Column-2: Contains number of ways of selecting three subsets of \(X\) according to column-1.Column-3: Contains conditional probabilities \(P\!\left(\dfrac{E}{E_1}\right)\) or \(P\!\left(\dfrac{E}{E_2}\right)\) where\(E\): Selecting three subsets of \(X\) according to column-1\(E_1\): Selecting three subsets of \(X\) such that \(n(A \cap B) = 5\)\(E_2\): Selecting three subsets of \(X\) such that \(n(A \cup B) = 5\).Column-1 Column-2 Column-3(I) \(A \cap B \cap C \supseteq \{2,3,4,5,6\}\) and \(A = B = C\) (i) 32 (P) \(P\!\left(\dfrac{E}{E_1}\right) = 0\)(II) \(A \cup B \cup C = \{3,4,5\}\) (ii) 242 (Q) \(P\!\left(\dfrac{E}{E_1}\right) = \dfrac{1}{{}^{10}C_5 \cdot 12^5}\)(III) \(A \cap B \cap C = \{3,4,5,6,7\}\) and \(A = B \neq C\) (iii) 243 (R) \(P\!\left(\dfrac{E}{E_2}\right) = \dfrac{31}{{}^{10}C_5 \cdot 12^5}\)(IV) \(A \cup B \cup C = \{6,7,8,9,10\}\) and \(A = B \neq C\) (iv) 343 (S) \(P\!\left(\dfrac{E}{E_2}\right) = 0\)[Note: \(S \supseteq T\) denotes \(S\) is a superset of \(T\), means \(S\) contains at least all elements of \(T\).]Which of the following options is the only correct combination?
Four numbers are chosen at random (without replacement) from the set \(\{1, 2, 3, \ldots, 20\}\).Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is \(1/85\).Statement-2: If the four chosen numbers from an AP, then the set of all possible values of common difference is \(\{\pm1, \pm2, \pm3, \pm4, \pm5\}\).
In a test, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices, only one answer being correct. The probability that he makes a guess is \(\dfrac{1}{3}\) and the probability that he copies the answer is \(\dfrac{1}{6}\). The probability that his answer is correct, given that he copies it, is \(\dfrac{1}{8}\). Find the probability that he knew the answer to the question, given that he correctly answers.
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 multiple of 4, is