Developed a simulation to determine the probability that 5 of 8 selected grade-12 students are studying English if the following is true.Of the 60 grade-12 students at a school, 45 are taking...

Developed a simulation to determine the probability that 5 of 8 selected grade-12 students are studying English if the following is true.

Of the 60 grade-12 students at a school, 45 are taking English. Suppose that 8 grade-12 students are selected at random for a survey.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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There are 60 students in the 12th grade

45 are taking English

We selected 8 students randomly :

We need to determine the probability that 5 out of the 8 students selected are studying English.

Since this is an independent event, then each selected student has the probability of 45/60 = 3/4

==> P (E) = 3/4

Then the probability of not studying English is:

P ( not E) = 1/4

Then P ( 5 are taking English out of 8) = P( 5 taking E)* P(3 not taking E)

= P (E) ^5 * P(not E) ^3

= (3/4)^ 5  * (1/4)^3

=  3^5 / 4^8

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neela | High School Teacher | (Level 3) Valedictorian

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The  random selection of students implies an equal chance of anybody being selected. Here we know that in the entire population of  60 stusents there are 45 english takers. So in a random section of 8 students , we again make a choice of 5 students.

In order that we should have all 5 english takers of the 5 students from 8 randomly selected students, there should be at least 5 english takers among  the eight chosen students. This event is possible if there are 5 english takers , or 6 english takers , or 7 english takers , or 8 english takers each with probability as below:

P(5 english takers in 8) =  8C5 * (p^5)(q^3).

P(6 english takers in 8) = 8C6*(p^6)(q^2).

P(7 english takers in 8) = 8C6*(p^7)(q^1)

P(all english takers in 8) = P^8.

All the above cases are excusive to each other.

So if there are 5 english takers, probabilty getting all 5 english takers in  a random choice from the 8  students is  (5C5/8C5). So the required probabilty is  {8C5*p^5*q^3}{5C5/8C5)} = p^5q^3.

Similarly probabilty of getting 5 english takers from 8 in which there are 6 english takers is {8C6*p^6*q^2}{ 6C5/8C5} = 3*p^6*q^2

Similarly getting a 5 english takers from an 8 radomly selected students  in which 7 there are english 7 is {8C7*p^7q}(7C5/(8C5) = 3p^7q^2

Similarly probabilty of getting a 5 english takers from 5 among 8 students in which all 8 are english takers is :

{8C8*P^8}(8C5/8C5)  = p^8.

Being exclusive events, the probablity of 5 english takers from 8 = P^5*q^3 +3p^6q^2+3p^7q+p^8 .

= p^5{q^3+3pq^2+ 3p^2q+p^3}.

= p^5{q+p)^3.

= p^5,  as q+p = 1.

= (3/4)^5.

Therefore , the probabity that among  the seleted 5 students out a random selection of 8 students , there are 5 english takers is  (3/4)^5.

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