Question

A box contains four black pieces of cloth, two striped pieces, and six dotted pieces. A piece is selected randomly and then placed back in the box. A second piece is selected randomly. What is the probability that: (a) both pieces are dotted? (b) the first piece is black and the second piece is dotted? (c) one piece is black and one piece is striped?

Answer 1

Answer:

a) ⇒$\frac{\textup{1}}{\textup{4}}$

b) ⇒  $\frac{\textup{1}}{\textup{6}}$

c) ⇒  $\frac{\textup{1}}{\textup{18}}$

Step-by-step explanation:

Data provided in the question:

Total Number of pieces = 4 + 2 + 6 = 12

P( Black piece )  = $\frac{\textup{4}}{\textup{12}}$

P( Striped piece ) = $\frac{\textup{2}}{\textup{12}}$

P( Dotted piece ) = $\frac{\textup{6}}{\textup{12}}$

Now,

a) P(Both the pieces are dotted) = P( Dotted piece ) × P( Dotted piece )

⇒ $\frac{\textup{6}}{\textup{12}}$  × $\frac{\textup{6}}{\textup{12}}$  

⇒ $\frac{\textup{36}}{\textup{144}}$  

⇒$\frac{\textup{1}}{\textup{4}}$

b) P(the first piece is black and the second piece is dotted)

= P( Black piece ) × P( Dotted piece )

⇒  $\frac{\textup{4}}{\textup{12}}$  × $\frac{\textup{6}}{\textup{12}}$  

⇒  $\frac{\textup{24}}{\textup{144}}$

⇒  $\frac{\textup{1}}{\textup{6}}$

c) P(one piece is black and one piece is striped)

= P( Black piece ) × P( Striped piece )

⇒ $\frac{\textup{4}}{\textup{12}}$  × $\frac{\textup{2}}{\textup{12}}$  

⇒  $\frac{\textup{8}}{\textup{144}}$

⇒  $\frac{\textup{1}}{\textup{18}}$