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2019 WAEC Mathematics Theory Three red balls, five green balls, and a number of blue balls are put together...

Mathematics
WAEC 2019

Three red balls, five green balls, and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \(\frac{1}{6}\) find;

(a) The number of blue balls in the sack 

(b) the probability of picking a green ball

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Explanation

(a) Let number of blue balls = n

Total number of balls = 3 + 5 + n 

= 8 + n

p(red ball) = \(\frac{\text{no. of red balls}}{\text{Total no. of balls}}\) 

\(\frac{1}{6} = \frac{3}{8 + n}\)

8 + n = 18 

n + 18 - 8 = 10

number of blue balls = 10

 

 

(b) P(green ball) = \(\frac{\text{No. of green balls}}{\text{Total no. of balls}}\)

= \(\frac{5}{3 + 5 + 10}\)

= \(\frac{5}{18}\)

= 0.2778


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WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995