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Suppose that a school band contains 10 students from the freshman class, 20 students from the sophomore class, 30 students from the junior class, and 40 students from the senior class. If 15 students are selected at random from the band, what is the probability that at least one student will be selected from each of the four classes Hint: First determine the probability that at least one of the four classes will not be represented in the selection.

Short Answer

Expert verified

The probability that at least one student will be selected from each of the four classes is \(0.7900\).

Step by step solution

01

Given information

A school band contains:

10 students from the freshman class.

20 students from sophomore

30 students from the junior class.

40 students from the senior class.

15 students are selected from the class randomly.

02

Describe the probability

The probability that at least one student will be selected from each of the four classes is expressed as,

\(\begin{array}{c}{\bf{P}}\left( {\bf{E}} \right){\bf{ = 1 - P}}\left( {{\bf{The}}\;{\bf{band}}\;{\bf{does}}\;{\bf{not}}\;{\bf{contain}}\;{\bf{any}}\;{\bf{one}}\;{\bf{class}}\;{\bf{member}}} \right)\\{\bf{ = 1 - P}}\left( {{{\bf{E}}^{\bf{c}}}} \right)\end{array}\)

The favourable outcomes that a freshman is not selected is,\(^{100 - 10}{C_{15}}{ = ^{90}}{C_{15}}\).

The favourable outcomes that a sophomore is not selected is,\(^{80}{C_{15}}\).

The favourable outcomes that a junior is not selected is,\(^{70}{C_{15}}\).

The favourable outcomes that a senior is not selected is,\(^{60}{C_{15}}\).

Total number of 15 band members from 100 students is,\(^{100}{C_{15}}\).

03

Compute the probability

The probability that student is not selected from exactly one class,

\(\begin{aligned}{}P\left( {{E^c}} \right) &= \frac{{^{90}{C_{15}}{ + ^{80}}{C_{15}}{ + ^{70}}{C_{15}}{ + ^{ + 60}}{C_{15}}}}{{^{100}{C_{15}}}}\\ &= 0.2100\end{aligned}\)\(\)

Thus, the probability that at least one student will be selected from each of the four classes is,

\(\begin{aligned}{}P\left( E \right) &= 1 - 0.2100\\ &= 0.7900\end{aligned}\)

Thus, the required probability is 0.7900.

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