Ahigh school theater club has 40 students, of whom 6 are left-handed. two students from the club will be selected at random, one at a time without replacement. what is the probability that the 2 students selected will both be left-handed?

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.  

The size of the sample space is the total number of possible outcomes  

The event space is the number of outcomes in the event you are interested in.  

so  

Let

x------> size of the event space

y-----> size of the sample space  

so

Find out the probability that the first student selected will be left-handed

we have

substitute

Simplify

Find out the probability that the second student selected will be left-handed

we have

substitute

Find out the probability that the 2 students selected will both be left-handed

Multiply the probabilities

Convert to percentage

Multiply by 100

So the Probability of choosing 2 left handed students = 1.92%

Probability is defined as  n (Favorable cases) /n ( Total sample space)

Total Number of students in the theater club  = 40

Total number of Left handed students = 6

So the probability of choosing one left handed student = 6/40 =3/20

Given that two students from the total students will be selected at random without replacement; which means that the sample size is decreased by 1.

So the new sample space is having 39 students only

Also one left handed student is already chosen without replacement hence only 5 left handed students are left into the sample space.

Hence , probability of choosing second left handed student = 5/39

So the Probability of choosing 2 left handed students from the sample space =

Probability of Choosing one Student Probability of Choosing another Student=  3/205/39

=0.0192 100 = 1.92%

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