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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

Best Answer 

 #1
avatar+17744 
+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
avatar+17744 
+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
avatar+92781 
+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 ) =      $$\frac{3}{4}+\frac{1}{3}-\frac{1}{4}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$

-----------------

 

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

 

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

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