There are 13 red balls, 6 green balls and 6 blue balls in a bag. What is the probability of drawing (i) red ball (ii) blue ball or a green ball? [2 MARKS]

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Solution

Sub-parts : 1 Mark each

(i) P (Red) = $\frac{13}{(13 + 6 + 6)}$ = $\frac{13}{25}$ = 0.52.

(ii) P (Blue or Green) = $\frac{6}{25} + \frac{6}{25}$ = $\frac{12}{25}$ = 0.48.

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There are 13 red balls, 6 green balls and 6 blue balls in a bag. What is the probability of drawing (i) red ball (ii) blue ball or a green ball? [2 MARKS]

Sub-parts : 1 Mark each (i) P (Red) = 13(13+6+6) = 1325 = 0.52. (ii) P (Blue or Green) = 625+625 = 1225 = 0.48.

Sub-parts : 1 Mark each

(i) P (Red) = $\frac{13}{(13 + 6 + 6)}$ = $\frac{13}{25}$ = 0.52.

(ii) P (Blue or Green) = $\frac{6}{25} + \frac{6}{25}$ = $\frac{12}{25}$ = 0.48.