Question

A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. Find the probability that a marble drawn is
(i) white
(ii) white and odd numbered
(iii) even numbered
(iv) red or even numbered.

Answer 1

Total number of marbles = (6 + 4) = 10
Let S be the sample space.
Then n(S) = number of ways of selecting one marble out of 10 = ¹⁰C₁ = 10 ways

(i)
Let E₁ = event of getting a white marble
∴ n(E₁) = ⁴C₁ = 4
Hence, required probability = $\frac{{}^{4}C_1}{{}^{10}C_1}=\frac{4}{10}=\frac{2}{5}$

(ii)
Let E₂ = event of getting a white marble, which is odd numbered.
i.e. E₂ = {13, 15}
∴ n(E₂) = 2
Hence, required probability = $\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{2}{10}=\frac{1}{5}$

(iii)
Let E₃ = event of getting an even numbered marble
i.e. E₃ = {2, 4, 6, 12, 14}
∴ n(E₃) = 5
Hence, required probability = $\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{5}{10}=\frac{1}{2}$

(iv)
Let E₄ = event of getting a red marble
i.e. E₄ = {1, 2, 3, 4, 5, 6}
∴ n(E₄) = 6
Now, P(E₄) = $\frac{6}{10}=\frac{3}{5}$ ...(i)
Let E₅ = event of getting even numbered marble
Then E₅ = {2, 4, 6, 12, 14}
i.e.n(E₅) = 5
Now, P(E₅) = $\frac{5}{10}=\frac{1}{2}$ ...(ii)
From (i) and (ii), we get:
E₄ ∩ E₅ = {2, 4, 6}
$\Rightarrow$n(E₄ ∩ E₅) = 3
$\Rightarrow$P(E₄ ∩ E₅) = $\frac{3}{10}$
By addition theorem, we have:
P (E₄ ∪ E₅) = P(E₄) + P (E₅) − P (E₄ ∩ E₅)
⇒ P (E₄ ∪ E₅) = $\frac{3}{5}+\frac{1}{2}-\frac{3}{10}=\frac{8}{10}=\frac{4}{5}$
Hence, required probability = P(E₄ ∪ E₅) = $\frac{4}{5}$