Solution : Here, n = 5 and r \(\ge\) 1 \(\therefore\) p(X = r) = \(^nC_r\) \(p^{n-r}\) \(q^r\) P(X \(\ge\) 1) = 1 – P(X = 0) = 1 – \(^5C_0 . p^5 . q^0\) \(\ge\) \(31\over 32\) [Given] \(\implies\) \(p^5\) \(\le\) 1 – \(31\over 32\) = \(1\over 32\) \(\therefore\) p \(\le\) \(1\over 2\) and […]
Solution : Given P(A \(\cup\) B)’ = 1/6, P(A \(\cap\) B) = 1/4 and P(A)’ = 1/4 \(\therefore\) P(A \(\cup\) B) = 1 – P(A \(\cup\) B)’ = 1 – \(1\over 6\) = \(5\over 6\) and P(A) = 1 – P(A)’ = 1 – \(1\over 4\) = \(3\over 4\) P(A \(\cup\) B) = P(A) +
Solution : Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2 \(\therefore\) P(A) + P(B) + P(C) = 1 then events A, B, C are exhaustive. If P(E) = Probability of introducing a new product, then as given P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5 = 0.5 \(\times\) 0.7 +
Solution : \(E_1\) : Event that first drawn ball is red, second is blue and so on. \(E_2\) : Event that first drawn ball is blue, second is red and so on. \(\therefore\) P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and \(\therefore\) P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\)
Solution : Total number of ways of dividing 48 cards(Excluding 4 Aces) in 4 groups = \(48!\over (12!)^4 4!\) Now, distribute exactly one Ace to each group of 12 cards. Total number of ways = \(48!\over (12!)^4 4!\) \(\times\) 4! Now, distribute these groups of cards among four players = \(48!\over (12!)^4 4!\) \(\times\) 4!4!
Solution : We know that probability of event = no. of favorable outcome/total outcome Sample space of event : { (H , H , H) , (H , H , T) , (H , T , H) , (H , T , T) , (T , H , H) , (T , H , T)
