Question

A bag contains 6 Cherry Starbursts and 24 other flavored Starburts. 5 Starbursts are chosen randomly without replacement.

Answer 1

Answer:

Required probability $=\frac{920}{23751}$

Step-by-step explanation:

Given that a bag contains 6 Cherry Starbursts and 24 other flavored Starbursts and 5 Starbursts are chosen randomly without replacement.  We need to find probability that 3 of the Starbursts drawn are cherry.

The number of ways of selecting 5 starbursts out of which 3 are cherry = 3 from 6 Cherry Starbursts 2 from 24 other flavored Starbursts

$=\begin{pmatrix}6\\ 3\end{pmatrix}\times \begin{pmatrix}24\\ 2\end{pmatrix}$

$=\frac{6!}{3!\times 3!}\times \frac{24!}{2!\times 22!}$

$=\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1}\times \frac{24\times 23\times 22!}{2\times 1\times 22!}$

$=5520$

Number of ways of selecting 5 starbursts out of 30 starbursts

= $\begin{pmatrix}30\\ 5\end{pmatrix}$

$=\frac{30!}{5!\times 25!}$

$=\frac{30\times 29\times 28\times 27\times 26\times 25!}{2\times 1\times 25!}$

$= 142506$

Required probability = $=\frac{favorable\, cases}{possible \, cases}$$=\frac{5520}{142506}=\frac{920}{23751}$