Question

In a statistics class of 42 students, 14 have volunteered for community service in the past. If two students are selected at random from this class, what is the probability that both of them have volunteered for community service? Round your answer to four decimal places. P(both students have volunteered for community service) the absolute tolerance is __________

Answer 1

Answer:

0.1056 is the probability that both the selected students volunteer for community service.                                                      

Step-by-step explanation:

We are given the following in the question:

Total number of students, n = 42

Number of students volunteering for community service, x = 14

Ways of selecting r objects from n objects is given by:

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

Ways of selecting two students from class =

$\binom{42}{2} = \dfrac{42!}{2!(42-2)!} = 861$

Ways of selecting 2 students volunteering for community service =

$\binom{14}{2} = \dfrac{14!}{2!(14-2)!} = 91$

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

P(both students have volunteered for community service)

$= \dfrac{91}{861} = 0.1056$

0.1056 is the probability that both the selected students volunteer for community service.

Answer 2

The correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

The calculation of the probability of both the students who have volunteered for the community service getting selected is shown by doing multiple calculations as under.

• It is assumed that the number of students is denoted by n. So, n=42.

• It is assumed that the students who have volunteered is denoted by x. So x= 14.

• Calculating further,

• The random 2 students can be selected by using the formula below and applying the given info to the formula we get,

• $\left \ ( {{n} \atop {r}} \right. )= \dfrac {n!}{2(42-2)!}\\\\\\\left \ ( {{42} \atop {2}} \right. )=861$

• Selecting 2 students out of the 14 volunteered by using the similar formula,

• $\left \ ( {{14} \atop {2}} \right. )= \dfrac{14!}{2(14-2)!}\\\\\\\left \ ( {{14} \atop {2}} \right. )= 91$

• Now calculating the probability by dividing the values derived from the above calculations,

• $\rm Probability= \dfrac{Favorable\ Observations}{Total\ Observations}\\\\\\\rm Probability= \dfrac{91}{861}\\\\\\\rm Probability= 0.1056$

• So we know that the probability of two students getting selected from the number of students who have volunteered for the community service is 0.1056.

Hence, correct statement is that the probability of both the students who have volunteered for community service the absolute tolerance will be 0.1056.

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