So the probability that at least one event occurs is 0.748.

[schinb65]

Two independent events have probabilities 0.1 and 0.3. What is the probability that at least one of
the events occurs?

I have an answer of .37, when I looked up the solution it is the same value but it was solved another way. I was just wondering if my logic would work.

Find $P(a \cup b) = P[a] + P - P[a \cap b]$
So I have P[a] and P, and I can find $P[a \cap b]$ from independence I can say that;
$P[a \cap b]$ = P[a]*P, correct?

Or is the only way that I can solve it which is easier then my approach,
1-P[neither event]
1-P(1-a)*P(1-b)
But I do not feel that this would be the first way that would pop into my mind.

 

[soroban]

Re: independent event

Hello, schinb65!

Two independent events have probabilities 0.1 and 0.3.
What is the probability that at least one of the events occurs?

I have an answer of 0.37.
When I looked up the solution, it is the same value,
but it was solved another way.
I was just wondering if my logic would work.

Find $P(a \cup b) = P[a] + P - P[a \cap b]$
So I have $P[a]$ and $P$, and I can find $P[a \cap b]$.
From independence I can say that: $P[a \cap b] = P[a]\!\cdot\!P$
Correct? . Yes!

The other approach is:
$1-P[\text{neither event}] \:=\:1 - P(a')\!\cdot\! P(b')$

But I do not feel that this would be the first way
that would pop into my mind.

If there were more events,
. . I'm sure this method would occur to you.Given: .\begin{Bmatrix}P[a] = 0.1 \\ P<b> = 0.3 \\ P[c] = 0.6 \end{Bmatrix}</b>

Find the probability that at least one event occurs.P[a\cup b \cup c] \:=\:1 - P[\sim\!a]\!\cdot\!P[\sim\!b]\!\cdot\!P[\sim\!c]

. . . . . . . . . =\:1 - (0.9)(0.7)(0.4)

. . . . . . . . . =\:1 - 0.252

. . . . . . . . . =\:0.748