In three independent flips of a coin where there is a 55% chance of flipping a tail, let A denote {first flip is a tail}, B denote {second flip is a tail}, C denote {first two flips are tails}, and D denote {three tails on the first three flips}. Find the probabilities of A, B, C, and D, and determine which, if any, pairs of these events are independent.

Pr(A) =  0.55

Pr (B) = 0.55

Pr (C) = 0.3025

Pr (D) = 0.166375

Step-by-step explanation:

Given that:

Pr(tail) = 55% = 0.55

Pr (head) = 1 - Pr(tails)

Pr (head) = 1 - 0.55

Pr (head) = 0.45

Pr(A) = first flip is a tail = 0.55

Pr (B) = second flip is a tail

Pr(B) = Pr(head on the  first flip and tail on the second flip) + P(tail on both flips)

Pr (B) = (0.45 × 0.55) + ( 0.55  × 0.55)

Pr (B) = 0.2475 + 0.3025

Pr (B) = 0.55

Pr (C) = 0.55 x 0.55

= 0.3025

Pr (D) = 0.55 x 0.55 x 0.55

= 0.166375

From above; we can conclude that Pr (A) and Pr (B) are independent.

A= 0.55,B=0.55,C=0.3025,D=0.1664; A and B are independent

Step-by-step explanation:

probability of tail:0.55

probability of not a tail: 0.45

A= P(tail and tail and tail)+P(tail and not tail and tail) + P(tail and not tail and not tail)+P(tail and tail and not tail)

A= 0.55×0.55×0.55+0.55×0.45×0.55+0.55×0.45×0.45+0.55×0.55×0.45

A=0.55

B= P(tail and tail and tail)+P(not tail and tail and tail)+P(tail and tail and not tail)+P(not tail and tail and not tail)

B=0.55×0.55×0.55+0.45×0.55×0.55+0.55×0.55×0.45+0.45×0.55×0.45

B= 0.55

C=P(tail and tail and tail)+P(tail and tail and not tail)

C=0.55×0.55×0.55+0.55×0.55×0.45

C=0.3025

D=P(tail and tail and tail)

D=0.55×0.55×0.55

D=0.1664

For events to be independent, the probability of occurance of one event must not be dependent on the outcome of other event.

Here, first and second flip have to be tail for C and D to happen. However, if first flip is a tail then second slip may or may not be a tail!

ian gonna lie that is way to much