Probability question 4

Question 4

Event A – possible equally probable outcomes: 0 and 1

Event B – possible equally probable outcomes: 0 and 1

Event A and event B each occurred once. The outcome of one of the two events was a 0 and occurred during a time interval t within a time interval T. The other event occurred during the time interval T. What is the probability that the outcome of the other event was also a 0?

Use the result to confirm the solution to problem 2.

What happens as t tends to 0? What if t = T? What is the range of this probability?

Answer 4

Independently of whether event A occurs before, after or simultaneously to event B, the only relevant combinations of these outcomes are:

  • Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T) AND Event B – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T)
  • Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T) AND Event B – outcome = 0, during time interval T but not during time interval t (probability = (1/2)x(T-t)/T) = (T-t)/(2T)
  • Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T) AND Event B – outcome = 1, during time interval T (probability = 1/2)
  • Event A – outcome = 0, during time interval T but not during time interval t (probability = (1/2)x(T-t/T) = (T-t)/2T)– AND Event B – outcome = 0 during time interval t (probability = (1/2)x(t/T) = t/2T)
  • Event A – outcome = 1 at any time in time interval T (probability = 1/2) AND Event B – outcome = 0 during time interval t (probability = (1/2)x(t/T) = t/(2T))

The probabilities of each of these combinations are:

Probability of combination 1) = (t/2T)(t/2T) = t2/4T2

Probability of combination 2) = (t/2T)((T-t)/2T) = t(T-t)/4T2

Probability of combination 3) = (t/2T)(1/2) = t/4T

Probability of combination 4) = ((T-t)/(2T))(t/2T) = t(T-t)/4T2

Probability of combination 5) = (1/2)( t/2T) = t/4T

Combinations 1), 2) and 4) are formed by two 0s, hence the probability of the second outcome being a 0 is:

(t2/4T2+t(T-t)/4T2+t(T-t)/4T2) / (t2/4T2+t(T-t)/4T2+t/4T+t(T-t)/4T2+t/4T) = ((2tT-t2)/4T2) / ((4tT-t2)/ 4T2) =

(2T-t) / (4T-t)

In problem 2 the outcomes were: boy or girl. So, 0 corresponds to a boy and 1 to a girl. One of the children was a boy born on a Tuesday, hence t = 1 day. All of the different days of the week occur over the course of a 7 day period, and the other child must have been born on one of the 7 days of the week; thus T = 7. Substituting these values we get:

(2×7-1) / (4×7-1) = 13/27.

As t tends to 0, the probability tends to: 2T/4T = ½.

If t = T, the probability becomes: (2T-T)/(4T-T) = T/(3T) = 1/3.

Therefore this probability must always be greater than (or equal to if we allow the events to occur in zero time) ½ and less than or equal to 1/3.

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