Suppose A and B are two events with P(A) = 0.6, P(B) = 0.4 P(AB) = 0.2. Find the

Question

Suppose A and B are two events with P(A) = 0.6, P(B) = 0.4 P(AB) = 0.2. Find the probabilities i) A Union B. (ii) A^c (iii) B^c, (iv) A^c B, (v) A Union B^c, (vi) A^c B^c. Events F, G and H are such that P(F) = 0.7, P(G) = 0.6, P(H) = 0.5. P(FG) = 0.4 P(FH) = 0.3. P(GH) = 0.2 P(FGH) = 0.1. Find (a) P(F Union G), (b), P(F Union G Union H), (c) P(F^c G^c H). A number is selected at random from the set {1, 2, 3, .. 1000}. What Is the probability that it is not divisible by 4, 7 or 9? (a) Suppose that in a state, license plates have three letters followed by three numbers, in a way that no letter or number is repeated in a single plate. Determine the number of possible license plates for this state. (b) A library has 800,000 books, and the librarian wants to encode by using a code word consisting of three letters followed by two numbers. Are there enough code words to encode all of these books with different code words? (a) In tossing four fair dice, what is probability of tossing, at most, one 4? (b) The elevator of a four-floor building leaves the first floor with six passengers and stops at all of the remaining three floors. If it is equally likely that a passenger gets off at any of these three floors, what is the probability that, at each stop of the elevator, at least one passenger departs? (a) Six fair dice are tossed. What Is the probability that at least two of them show the same face? (b) Five boys and five girls sit at random in a row. What is the probability that the boys are together and the girls are together?

Explanation / Answer

Soluion:-

1. P(A) = 0.6
P(B) = 0.4
P(AB)= 0.2
i) P(A or B) = P(A) + P(B) - P(AB)
= 0.6 + 0.4 - 0.2
= 0.8
ii) P(Ac) = 1 - P(A)
= 1 - 0.6 = 0.4
iii) P(Bc) = 1 - P(B)
= 1 - 0.4 = 0.6
iv) P(AcB) = P(Ac) * P(B)
= 0.4 * 0.4
= 0.16
v) P(A or Bc) = P(A) + P(Bc) - P(ABc)
= 0.6 + 0.6 - 0.6*0.6
= 0.84
vi) P(AcBc) = P(Ac) * P(Bc)
= 0.4 * 0.6
= 0.24

2. P(F) = 0.7, P(G) = 0.6 and P(H) = 0.5
P(FG) = 0.4, P(FH) = 0.3, P(GH) = 0.2 and P(FGH) = 0.1
i) P(F or G) = P(F) + P(G) - P(FG)
= 0.7 + 0.6 - 0.4
= 0.9
ii) P(F or G or H) = P(F) + P(G) + P(H) - P(FG) - P(FH) - P(GH) + P(FGH)
= 0.7 + 0.6 + 0.5 - 0.4 - 0.3 - 0.2 + 0.1
= 1
iii) P(FcGcH) = P(Fc) * P(Gc) * P(H)
= 0.3 * 0.4 * 0.5
= 0.06

3.

=(1000/4)+(1000/7)+(1000/9)-(1000/28)-(1000/36)-(1000/63)+(1000/252) = 428.57

p = 0.42857

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