Question

Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.

Answer 1

Let X denote the number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.
Now,

\[P\left( X = 0 \right)\]
\[ = P\left( \text{ no defective mango } \right)\]
\[ = \frac{{}^{15} C_4}{{}^{20} C_4}\]
\[ = \frac{1365}{4845}\]
\[ = \frac{91}{323}\]
\[P\left( X = 1 \right)\]
\[ = P\left( 1 \text{ defective mango } \right)\]
\[ = \frac{{}^5 C_1 \times^{15} C_3}{{}^{20} C_4}\]
\[ = \frac{2275}{4845}\]
\[ = \frac{455}{969}\]
\[P\left( X = 2 \right)\]
\[ = P\left( 2 \text{ defective mangoes } \right)\]
\[ = \frac{{}^5 C_2 \times^{15} C_2}{{}^{20} C_4}\]
\[ = \frac{1050}{4845}\]
\[ = \frac{70}{323}\]
\[P\left( X = 2 \right)\]
\[ = P\left( 3 \text{ defective mangoes } \right)\]
\[ = \frac{{}^5 C_3 \times^{15} C_1}{{}^{20} C_4}\]
\[ = \frac{150}{4845}\]
\[ = \frac{10}{323}\]
\[P\left( X = 3 \right)\]
\[ = P\left( 4 \text{ defective mangoes }  \right)\]
\[ = \frac{{}^5 C_4}{{}^{20} C_4}\]
\[ = \frac{5}{4845}\]
\[ = \frac{1}{969}\]

Thus, the probability distribution of X is given by

x	P(X)
0	\[\frac{91}{323}\]
1	\[\frac{455}{969}\]
2	\[\frac{70}{323}\]
3	\[\frac{10}{323}\]
4	\[\frac{1}{969}\]