1. ## Probility questions:

doing some probability questions for a assignment, I think i have them right but maths isnt my strong suit could anyone look at these two questions comment if these are correct

A computer randomly chooses 4 different letters from the 26 letters of the alphabet.
Calculate the probability that the chosen letters spell the word ZONE

solution:
26*26*26*26 =456976
p(zone) is 1 over 456976

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Suppose A and B are events with P(A) = 0.4 and P(B) = 0.4.
Find the probability that
(i) A does not occur
(ii) B does not occur
(iii) A or B occurs
(iv) Neither A nor B occurs

Solution:
(i) P(not A)= 1-0.4 = 0.6 = 3 over 5
(ii) P(not B) = 1 - 0.4 = 0.6 = 3 over 5
(iii)P(AorB) = 0.4 +0.4 =0.8 = 4 over 5
(iv) P(not A or B) = 1 - 0.8 =0.2 = 1 over 5 (*sorry i dont know hoe to edit this to show fractions)

2. ## Re: Probility questions:

Originally Posted by ronanbrowne88
doing some probability questions for a assignment, I think i have them right but maths isnt my strong suit could anyone look at these two questions comment if these are correct.

A computer randomly chooses 4 different letters from the 26 letters of the alphabet.
Calculate the probability that the chosen letters spell the word ZONE
solution:
[/B]26*26*26*26 =456976
p(zone) is 1 over 456976
Read the question carefully. It says 4 different letters. So it it would be $26\cdot 25\cdot 24 \cdot 23$.

Originally Posted by ronanbrowne88
Suppose A and B are events with P(A) = 0.4 and P(B) = 0.4.
Find the probability that
(i) A does not occur
(ii) B does not occur
(iii) A or B occurs
(iv) Neither A nor B occurs

Solution:
(i) P(not A)= 1-0.4 = 0.6 = 3 over 5
(ii) P(not B) = 1 - 0.4 = 0.6 = 3 over 5
(iii)P(AorB) = 0.4 +0.4 =0.8 = 4 over 5
(iv) P(not A or B) = 1 - 0.8 =0.2 = 1 over 5
You have answered parts (i) & (ii) correctly.

However the is no way possible to answer parts (iii) & (iv).
There is not enough information given.
We need to know if $A~\&~B$ are independent or else the value of $\mathcal{P}(A\cap B)$.

3. ## Re: Probility questions:

i think they are independent, if that is the case is the above correct

4. ## Re: Probility questions:

Originally Posted by ronanbrowne88
i think they are independent, if that is the case is the above correct
No. Part (iii) is $\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B).$