# probability - Difference between two statements

I have the following scenario.

There are a number of people on an island. Each person has a chance of 0.015 of dying on any given day, independent of other people on the island. I have two different probabilities for two statements that were calculated by the professor, but I can't see the difference between the meaning of the two statements.

Statement 1: Person A survived the sixth day.

Probability of statement 1:

$$P=(1-0.015)^5\cdot 0.015=0.0139$$ (probability person A died on the 6-th day)

$$P(Statement 1)=1-0.0139=0.986$$

Statement 2: Person A survived for 6 straight days.

Probability of statement 2:

$$P(Statement 2)=0.985^6=0.913$$

I understand how each probability was calculated, but I don't see the difference between each statement logically. Please explain what is the difference, or give an example when statement 1 occurs but statement 2 doesn't, or vice versa.

Thank you.

There is nothing in the model that prevents a person of dying on day one and then living on day two. This is why the two statements are different:

• Statement 1 is the negation of "the person did not die on days 1 through 5 and died on day 6"
• Statement 2 means "the person did not die on days 1 through 6"

Let $$D_i$$ be the event of the person dying on day $$i$$, then

$$\text{Statement 1} = \neg\big((\neg D_1)(\neg D_2)(\neg D_3)(\neg D_4)(\neg D_5) D_6\big) =$$ by De Morgan $$= D_1 \vee D_2 \vee D_3 \vee D_4\vee D_5\vee \neg D_6$$ which is true when the person dies on any one of the days 1 through 5.

• Ok, I understand now. Thanks! – Alexey Kiryushkin Feb 27 at 12:18

Given a different event (not dying and survivinf) they would be different. But since the event of surviving the 6th day induces surviving previous days, they have the same probability. If event i.e. was being healthy, probabilities would be different imo