What is the probability of exactly one tall student out of two students?

  • Thread starter neubreed
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In summary, the probability that exactly one out of two different students called on at random will be tall is 624/1225.
  • #1
neubreed
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Homework Statement


26 out of 50 total students are tall. If 2 different students are called on at random, what is the probability that exactly one is tall?

Homework Equations


please see below


The Attempt at a Solution


so here's what I ended up with:
1st student is tall: 26/50 x 24/49=312/1225
2nd student is tall: 24/50 x 26/49= 312/1225

I'm not sure whether I now multiply the two fractions or add them... Am I on the right track?

Thanks!
 
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  • #2
neubreed said:

Homework Statement


26 out of 50 total students are tall. If 2 different students are called on at random, what is the probability that exactly one is tall?

Homework Equations


please see below


The Attempt at a Solution


so here's what I ended up with:
1st student is tall: 26/50 x 24/49=312/1225
2nd student is tall: 24/50 x 26/49= 312/1225

I'm not sure whether I now multiply the two fractions or add them... Am I on the right track?

Thanks!

Yes, if you want to do it that way you are on exactly the right track. The two events you have are mutually exclusive. What do you think about the question of whether to add or multiply?
 
  • #3
I'm leaning towards adding, but the end result seems a little high... Would 624/1225 be the asnwer?
 
  • #4
Why is that "high"? It is just about 1/2 and just about 1/2 of the students are "tall".

If A and B are "equally likely" (probability of each 1/2) then "AA", "AB", "BA", and "BB" all have probability 1/4 so AB+ BA has probability 1/2.
 
  • #5
neubreed said:
I'm leaning towards adding, but the end result seems a little high... Would 624/1225 be the asnwer?

Yes, it is. You can check it using the combinations formula C(n,k) if you know that. There are C(26,1) ways to choose the tall, C(24,1) ways to choose the other and C(50,2) ways total ways to choose 2 students. C(26,1)*C(24,1)/C(50,2)=624/1225.
 

1. What does "probability of exactly 1" mean?

The "probability of exactly 1" refers to the likelihood of an event occurring exactly once in a series of repeated trials.

2. How is the probability of exactly 1 calculated?

The probability of exactly 1 can be calculated by dividing the number of desired outcomes (1) by the total number of possible outcomes.

3. Can the probability of exactly 1 be greater than 1?

No, the probability of an event occurring exactly once cannot be greater than 1. It is expressed as a decimal or percentage between 0 and 1.

4. How is the probability of exactly 1 different from the probability of at least 1?

The probability of exactly 1 refers to the specific outcome of an event occurring once, while the probability of at least 1 includes the possibility of the event occurring more than once.

5. How can the probability of exactly 1 be used in real-life scenarios?

The probability of exactly 1 can be used in various scenarios such as predicting the chances of winning a lottery or estimating the likelihood of a certain number of accidents in a given time period.

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