6 LED televisions and 4 LCD televisions are displayed in digital store

1.	6 LED televisions and 4 LCD televisions are displayed in digital store A whereas 5 LED televisions and 3 LCD televisions are displayed in digital store B. One of the two stores is randomly selected and one television is selected from that store. Find the probability that it is an LCD television.
Answer» Digital store A: 6 LED TV + 4 LCD TV = 10 TVs are displayed Digital store A: 6 LED TV + 4 LCD TV = 10 TVs are displayed A₁ = Event that a store is selected from the store A and B ∴ p(A₁) = \(\frac{1}{2}\) A₂ = Event that LCD TV is selected from the store A ∴ p(A₂) = \(\frac{^4C_1}{^{10} C_1} = \frac{4}{10}\) B₁ = Event that if store A is selected then LCD TV is selected. ∴ p(B₁) = p(A₁) ∙ p(A₂) = \(\frac{1}{2}×\frac{4}{10} = \frac{4}{20}\) B₂ = Event that if store B is selected then LCD TV is selected ∴ p(B₂) = p(A₁) ∙ p(A₃) = \(\frac{1}{2}×\frac{3}{8} = \frac{3}{16}\) C1 = Event that if one of the two stores is selected then LCD TV Is selected = B₁ ∪ B₂ ∴ P(C₁) = P(B₁) + P(B₂) = \(\frac{4}{20}+\frac{3}{16}\) = \(\frac{16+15}{80}\) = \(\frac{31}{80}\)

6 LED televisions and 4 LCD televisions are displayed in digital store A whereas 5 LED televisions and 3 LCD televisions are displayed in digital store B. One of the two stores is randomly selected and one television is selected from that store. Find the probability that it is an LCD television.

Answer»

Digital store A: 6 LED TV + 4 LCD
TV = 10 TVs are displayed

A₁ = Event that a store is selected from the store A and B
∴ p(A₁) = \(\frac{1}{2}\)

A₂ = Event that LCD TV is selected from the store A
∴ p(A₂) = \(\frac{^4C_1}{^{10} C_1} = \frac{4}{10}\)

B₁ = Event that if store A is selected then LCD TV is selected.
∴ p(B₁) = p(A₁) ∙ p(A₂) = \(\frac{1}{2}×\frac{4}{10} = \frac{4}{20}\)

B₂ = Event that if store B is selected then LCD TV is selected
∴ p(B₂) = p(A₁) ∙ p(A₃) = \(\frac{1}{2}×\frac{3}{8} = \frac{3}{16}\)

C1 = Event that if one of the two stores is selected then LCD TV Is selected

= B₁ ∪ B₂

∴ P(C₁) = P(B₁) + P(B₂)

= \(\frac{4}{20}+\frac{3}{16}\)

= \(\frac{16+15}{80}\)

= \(\frac{31}{80}\)