A point is randomly chosen in a square [0, 1] times [0, 1]. The event A is that

Question

A point is randomly chosen in a square [0, 1] times [0, 1]. The event A is that the point is in a square with vertices (0, 0), (0, l/2), (l/2, 1/2), (1/2, 0). The event B is that the point is in a square with vertices (1/4, 1/4), (1, 1/4), (1, 1), (1/4, 1). (See Fig. 1a). Find the probability^1 (a) P(A) = (b) P(B) = (c) P (AB) = (d) P(A or B) = (e) P(A^e) = (f) P(B^e) = (g) P(A^c and B^c) = (h) P(A or B)^c) = (i) P(A|B) = (j) P(B|A) = Are events A and B independent?_____ A point is randomly chosen in a square [0, 1] times [0, 1]. the event A is that the point is in a square with vertices (0, 0), (0, 1/2), (1/2, 1/2), (1/2, 0). The event B is that the

Explanation / Answer

P(A) = Probability that a selected point is in area of square/ Probability of the point in a square of area 1*1

Area of square A = side length * side length = 1/2*1/2=1/4

Area of randomly chosen sqare [0,1]*[0,1] = 1*1=1

So P(A)=area of square A/area of randomly chosen square = (1/4)/1 = 1/4 --> P(A)=1/4

Similarly,

Area of square B = side length * side length = 3/4*3/4=9/16

Area of randomly chosen sqare [0,1]*[0,1] = 1*1=1

P(B) = area of square B/area of randomly chosen square = (9/16)/1 = 9/16 --> P(B) = 9/16

P(AB) = probability that the point lies in both squares. i.e., the probability that the point lies in the intersection area of Square A and square B.

So in figure 1a. if you see the common intersection square, the side length of this square will be (1/2-1/4) = 1/4

Area of intersection for square A and B = sidelength * sidelength = (1/4)*(1/4) = (1/16)

P(AB) = Area of intersection for square A and B/area of randomly chosen square = (1/16)/1 = 1/16 --> P(AB) = 1/16

P(A or B) = P(A) + P(B) - P(AB) --> So from the above calculations,

P(A or B ) = (1/4) + (9/16)-(1/16) = 3/4 --> P(A or B) = 3/4

P(Ac) is defined as the probability that the chosen point is not in Square A but any where else in the randomly chosen square.

P(Ac) = 1 - P(A) --> P(Ac) = 1- (1/4) = 3/4--> P(Ac) = 3/4

Similarly, P(Bc) = 1 - P(B) --> P(Bc) = 1- (9/16) = 7/16--> P(Bc) = 7/16

P(Ac and Bc) = Probability that the chosen point does not lie in both squares A and B.

P(Ac and Bc) = 1- P(Aor B) = 1- (3/4) = 1/4

P((A or B)c) is same as P(Ac and Bc) i.e., the probability that the point does not lie in both the squares.

P((A or B)c) = P(Ac and Bc) = 1/4

P(A/B) is defined as the Probability that the point lies in A given that the point is in B

From fig 1a. it can be written as P(A/B) = (Probability that it lies in intersecting area of squares A and B)/ Probability that the point lies in square B = P(A and B)/ P(B) = (1/16)/(9/16) = (1/9) --> P(A/B) = 1/9

Similarly, P(B/A) = P(A and B)/ P(A) = (1/16)/(1/4) = (1/4) --> P(B/A) = 1/4

Are events A and B independent?

Ans: No A and B are not independent events as there is a common area between them.

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