Probability Questions

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)