Answer to Question #192106 in Statistics and Probability for Gabby

Question

• QUESTION FOUR [20] 4.1 A box contains balls, drawn at random. (The first ball is not replaced). Find the probability that both balls are of the same colour. (4) 5 red 6 white balls and 9 black balls. Two balls are 4.2 At the Durban beachfront of days. If rain it rains on 20% of days and it is windy on 30% and wind are independent, any particular day find the probability that: 4.2.1 it rains and is windy. (3) 4.2.2 it does not rain and is not windy. (3) 4.2.3 it rains or is windy. (3) 4.2.4 it does not rain or is not windy. (3) 4.3 Draw a Venn diagram to illustrate the data in question 4.2 above. (4)

Answer 1

So, our first question is

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at random. (The first ball is not replaced). Find the probability that both balls are of the same colour.

Ans:) Total balls $= 9+6+5 = 20$

Hence, Total number of ways of selecting two balls without replacement $= ^{20}C_2$

Therefore, the probability that both balls are of the same colour $= \dfrac{^5C_2+^6C_2+^9C_2}{^{20}C_2}$

Second question is:

At the Durban beachfront it rains on 20% of days and it is windy on 30% of days. If rain and wind are independent, any particular day find the probability that:

4.2.1 it rains and is windy.

4.2.2 it does not rain and is not windy.

4.2.3 it rains or is windy. 

4.2.4 it does not rain or is not windy.

Ans:) a.) $P(A)= 0.2$ and $P(B) = 0.3$

$P(A\hspace{2mm} and\hspace{2mm} B) = P(A).P(B) = (0.2).(0.3) = 0.06$

b.)$P(\bar A \hspace{2mm}and \hspace{2mm} \bar B) = (1-P(A))(1-P(B)) = (0.8).(0.7) = 0.56$

c.)$P(A\hspace{2mm} or\hspace{2mm} B) = P(A)+P(B)-0.2\times 0.3 = 0.44$

d.)$P(\bar A\hspace{2mm} or\hspace{2mm} \bar B) = \bar P(A)+ \bar P(B)-0.8 \times 0.7 = 0.94$

e.) Venn diagram: