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# Two letters are chosen at random from the word GEESE and two are chosen at random from the word PLEASE. What is the probability that all four letters (cont.)

are E's or none of the letters is an E. answer 3/50

I have the correct answer for these events in Adding Probabilities, but I am a substitute teacher, my degree is in music. I need to help the students to use the correct process to arrive at the answers.

Thank you in advance for any help you can give me.

### 3 Answers by Expert Tutors

Michael W. | Patient and Passionate Tutor for Math & Test PrepPatient and Passionate Tutor for Math & ...
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R,

GEESE:

Success is:  picking 2 e's in a row.

The probability of choosing a single E from GEESE is 3/5.
Once I've chosen an E, the probability of choosing a second E, from the remaining letters, is 2/4.  One less letter available, and one less E available.  That's 1/2.

The probability of doing both of those, in succession, is the first times the second.  (Think of it this way:  3/5 of the time, I pick an E, and then of that 3/5, I pick another E 1/2 of the time.  So, to get 1/2 of 3/5, I multiply.)

So, that's 3/5 * 1/2, which reduces to 3/10.

Now, let's do PLEASE the same way:

The probability of choosing a single E is 2/6 (or 1/3).
Once I've chosen an E, the probability of getting the remaining E, out of the remaining letters, is 1/5.
The probability of doing both of those, in succession, is 1/3 * 1/5, or 1/15.

Now, put together the geese and the please, the probability of getting 2 E's from geese AND the 2 E's from please:

3/10 * 1/15 = 3/150, or 1/50.

Again, 3/10 of the time, I get 2 E's from geese, and then of that 3/10, I get another 2 E's from please 1/15 of the time.  1/15 of 3/10 is 1/50.

Those are the situations where I get all 4 E's.  To do the "everything but E's" scenario, I do the same thing, but only choosing the non-E's.  The short version:

GEESE:  2/5 * 1/4 = 1/10
PLEASE:  4/6 * 3/5 = 2/5
No E's from both:  1/10 * 2/5 = 2/50

We now need all 4 E's or no E's.  If these are completely separate possibilities, then I can just add them together.  For an example of when they're not completely separate, think of a deck of cards.  If I'm trying to draw a ten or a spade, there's actually a card that works for both:  namely, the ten of spades.  So, if I just add the probabilities together (the four tens, plus the thirteen spades in the deck), I just counted the ten of spades twice, and that's going to mean that I've overcounted my possible success by a card.

In this situation, "All 4 E's" and "no E's" don't mix, so 1/50 + 2/50 = 3/50.

Hope this helps!  Maybe other tutors have a simpler way of explaining it, too...

Thank you.  That 's where I was headed with the problem as well.
Jon P. | Knowledgeable Math, Science, SAT, ACT tutor - Harvard honors gradKnowledgeable Math, Science, SAT, ACT tu...
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First look at GEESE.  There are 3 E's in 5 letters.  So the probability of picking an E on the first pick is 3/5.  Then, if you chose an E the first time, there are 4 letters to choose from of which 2 are E's.  So the probability of also picking an E will be 2/4.  So multiply 3/5 by 2/4 to get the probability of picking 2 E's from GEESE.  That's 3/10.

Follow the same logic in PLEASE and you get 2/6 * 1/5 = 1/15 as the probability of picking 2 E's from PLEASE.

Since you need the probability of picking 2 E's from BOTH words, you have to multiply the probabilities for each word together:  3/10 * 1/15 = 1/50.

What about the probability of picking NO E's?  Think of not picking as E as the same picking something other than E, a "non-E."  First look at GEESE.  There are two non-E's, G and S, so the probability of picking a non-E on the first pick is 2/5.  On the second pick, there would only be one non-E left out of 4 letters, for a probability of 1/4.  So the probability of not picking an E in both picks is 2/5 * 1/4 = 1/10.

Again, follow the same logic in PLEASE and you get 4/6 * 3/5 = 2/5 as the probability of not picking an E in both picks.  So in order to pick all non-E's in both words, you have to multiply the non-E probabilities together:  1/10 * 2/5 = 2/50.

Now the original question asked for the probability of picking all E's or no E's.  So you add the two probabilities we just calculated:  1/50 + 2/50 = 3/50.

I hope that explains it.  I know that working out problems like this can be a little obscure, especially without having been taught the basics first.

William F. | Math & Computer Science TutorMath & Computer Science Tutor
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Without order
6 x 2 = 12
20 x 30
12/600
6/300
3/150
1/50

2
4 x 3 = 12
2 x 12 = 24
24/600
2/50

The explanation is quite hairy, because of combinatorics and the use of sets; which is the only way I can explain this.

The probability that all four letters are 'E's will be the number of combinations of EE that can created from the three 'E's in GEESE, divided by the total number of all possible combinations of any two letters in GEESE. Do the same for PLEASE; & after some work, you get 1/50.

The probability that no letter is an 'E' will be the number of all two letter combinations (that don't include an 'E'!) in GEESE, divided by the total number of all possible combinations of any two letters in GEESE. Do the same for PLEASE; & after some work, you get 2/50.

Now there is an equation that states

P(AUB) = P(A) + P(B) - P(AB), where A and B are sets. Since the probabilities that we calculated involve mutually exclusive sets (or have no objects in common), we can just add the probabilities. So 1/50 + 2/50 = 3/50.

But I'm very sorry that I can't give a clearer explanation that won't involve sets and combinatorics.

Tutors, please feel free to review my argument.