Question

A bucket contains 10 balloons and 20 balls out of which 3 balloons and 5 balls are damaged. if we choose two items at a random, find the probability that either both are balls or both are not damaged

Answer 1

Correct answer is $P(A \cup B) = 0.726$

Explanation:

$\textrm{Sample space} = \textrm{20 balls} + \textrm{10 balloons}$

$= 30 things$

Number of balls which are damaged $= 5$

Number of balloons which are damaged $= 3$

Let A be the event of getting balls

$n(A) = \binom{20}{2} = \frac{20 \times 19}{2 \times 1}$

$P(A) =\frac{ n (A)}{n (S)}$

$P(A) = \frac{190}{\binom{30}{2}}$

$P(A) = \frac{190 \times 2}{30\times 29}$ $= \frac{38}{87}$

Let B be the event of getting undamaged

$n(B) =\textrm{Total number of things - number of damaged things}$

$(B) = 30 - 8 = 22$

$n(B) =\binom{22}{2} = \frac {22 \times 21}{2 \times 1}$

$P(B) = \frac{n(B)}{n(S)}$

$P(B) = \frac{11 \times 21 \times 2}{30 \times 29}$

$P(B) = \frac{77}{145}$

$n(A \cap B) = \binom{15}{2}$

$P(A \cap B) = \frac{n(A \cap B)}{n (S)}$

$P(A \cap B) = \frac{15 \times 14}{30 \times 29}$

$P(A \cap B) = \frac{14}{58}$

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$= \frac{38}{87} + \frac{77}{145} - \frac{14}{58}$

$=\frac{316}{435}$  

 Correct probability $P(A \cup B) = 0.726$