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

a) ⇒
`\frac{\textup{1}}{\textup{4}}`

b) ⇒
`\frac{\textup{1}}{\textup{6}}`

c) ⇒
`\frac{\textup{1}}{\textup{18}}`

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}}`