Question 1185015
<br>
There are 140 students, of whom 49 are mechanical engineering majors.<br>
The total number of ways of choosing 5 students is C(140,5); the number of ways of choosing 5 students from those that are not mechanical engineering majors is C((140-49),5) = C(91,5).<br>
The probability of choosing 5 students of whom none are mechanical engineering majors is<br>
{{{C(91,5)/C(140,5)}}}<br>
{{{((91*90*89*88*87)/(5*4*3*2*1))/((140*139*138*137*136)/(5*4*3*2*1)) = (91*90*89*88*87)/(140*139*138*137*136)}}} = 0.11153 to 5 decimal places.<br>
Note you would probably do the calculation using a calculator that can calculate numbers like C(140,5).<br>
Note you could also solve the problem by considering choosing the 5 students one at a time.<br>
the probability that the first student chosen is not a mechanical engineering major is 91/140;
the probability that the second student chosen is not a mechanical engineering major is 90/139;
the probability that the third student chosen is not a mechanical engineering major is 89/138;
the probability that the fourth student chosen is not a mechanical engineering major is 88/137;
the probability that the fifth student chosen is not a mechanical engineering major is 87/136<br>
The probability that none of the 5 is a mechanical engineering student is<br>
{{{(91/140)(90/139)*(89/138)*(88/137)*(87/136)}}}<br>
which is the same calculation as before.<br>