Mathematics: Abox 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?

80 beedsStep-by-step explanation:x = 4*20x = 80...

Abox contains four black pieces of cloth, two striped

EllaSue

08.01.2022

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 )

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

⇒ $\frac{\textup{8}}{\textup{144}}$

⇒ $\frac{\textup{1}}{\textup{18}}$

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