If the probability of spinning an A on spinner 1 is 3/4 and the probability of spinning E on spinner 2 is 1/3, what is the probability of spinning at least one vowel?

Guest Feb 2, 2015

#1**+10 **

Assuming that they only vowels are an A on spinner 1 and an E on spinner 2:

There are four possibilities:

spinner 1 = A and spinner 2 = E = (3/4) x (1/3) = ?

spinner 1 = A and spinner 2 = not E = (3/4) x (2/3) = ?

spinner 1 = not A and spinner 2 = E = (1/4) x (1/3) = ?

spinner 1 = not A and spinner 2 = not E = (1/4) x (2/3) = ??

If you add the answers to the three possibilities that have a '?', you will get the probaiblity of getting at least one vowel.

geno3141 Feb 2, 2015

#1**+10 **

Best Answer

Assuming that they only vowels are an A on spinner 1 and an E on spinner 2:

There are four possibilities:

spinner 1 = A and spinner 2 = E = (3/4) x (1/3) = ?

spinner 1 = A and spinner 2 = not E = (3/4) x (2/3) = ?

spinner 1 = not A and spinner 2 = E = (1/4) x (1/3) = ?

spinner 1 = not A and spinner 2 = not E = (1/4) x (2/3) = ??

If you add the answers to the three possibilities that have a '?', you will get the probaiblity of getting at least one vowel.

geno3141 Feb 2, 2015

#2**+5 **

Thanks Geno,

If the probability of spinning an A on spinner 1 is 3/4 and the probability of spinning E on spinner 2 is 1/3, what is the probability of spinning at least one vowel?

Geno's answer is really good and will work perfectly.

Here is another method

first I will assume that the only vowel on 1 is A so P(vowel on 1) = 3/4

and the only vowel on 2 is E so P(cowel on 2) = 1/3 (Geno made the same assumption)

I could add those together but if I did I would be counting the probability of getting 2 vowels twice

P(2 vowels) = 3/4*1/3 = 1/4

so

P(of spining at least one vowel ) = $$\frac{3}{4}+\frac{1}{3}-\frac{1}{4}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$

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$$\boxed{P(A or B) = P(A)+P(B)-P(AandB)}$$

This is the 2 spinner outcomes. So how if you just add them up the cross over bit will be counted twice?

Melody Feb 2, 2015