Probability of event given another event occurs twice in a row

[tmt1]

I have $P(B) = 0.4$ and $P(\lnot B) = 0.6$.

$P(TS|B) = 0.7$ and $P(TS|\lnot B) = 0.25$

$P(B|TS) = 0.65116$ and $P(\lnot B|TS) = 0.34884$ (from bayes theorem).

Now, if we get $B$ or $\lnot B$, and we get the same event twice in a row so we get $B$ then $B$ or $\lnot B$ then $\lnot B$, what is the probability of $TS$ given we get the same event twice in a row?

 

[jvicens]

Hello,

Thank you for sharing your probability values with us. Based on the information you provided, the probability of getting the same event twice in a row is 0.4*0.4 + 0.6*0.6 = 0.52. This means that in this scenario, there is a 52% chance of getting either B or ¬B twice in a row.

Now, to calculate the probability of TS given that we get the same event twice in a row, we can use the conditional probability formula:

P(TS|B∩B) = P(TS∩B)/P(B∩B)

= P(TS|B)*P(B)/P(B∩B)

= 0.7*0.4/0.4*0.4

= 0.7/0.4

= 1.75

Therefore, the probability of TS given that we get the same event twice in a row is 1.75 or 175%. This means that there is a high chance of getting TS in this scenario.

I hope this helps to answer your question. Let me know if you have any further inquiries.