Double spins and not mutually exclusive.

datadan

New member
Joined
Oct 19, 2010
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5
Hello,
Can someone please check my reasoning on this problem:

A fair spinner (with 20 slots) is spun twice. Find P(get a 4 OR get an number divisible by 4).
Let A = 4 divided the number

So by addition principle we get: P(4) + P(A) - p(4 and A)

We find P(4). We can get 4 on the first spin, on the second, OR on both
4 on first spin= (1/20)(19/20) =19/400
4 on second = (19/20)(1/20) =19/400
4 on both = (1/20)(1/20) =1/400

P(4) = 39/400

We find P(A). We can either have an A in the first spin or A in the second, or A on in both spins.
A on the first spin = (5/20)(15/20) = 75/400
A on the second spin = (15/20)(5/20)= 75/400
A on both spins = (5/20)(5/20) = 25/400

So P(A) = 175/400.

As these events are NOT mutually exlusive we need to account for the overlaps (4)

P(4) = 39/400 see above.

So P(4) + P(A) - P(4 and A) = 39/400 + 175/400 - 39/400 = 175/400 =43.75%

Is this reasoning correct?

Thank you,
 
Because probability is my weakest subject, and I want to make sure my reasoning is correct.

Thank you,
 
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