Q.8. A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table:

Outcome 1 2 3 4 5 6

Frequency 178 152 156 150 176 188

Find the probability of getting each outcome.

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Solution

Dear Student,

$T o t a l n u m b e r o f t h r o w s = 1000 P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 1 = \frac{F r e q u e n c y o f 1}{t o t a l t h r o w s} = \frac{178}{1000} = \frac{89}{500} P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 2 = \frac{F r e q u e n c y o f 2}{t o t a l t h r o w s} = \frac{152}{1000} = \frac{76}{500} P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 3 = \frac{F r e q u e n c y o f 3}{t o t a l t h r o w s} = \frac{156}{1000} = \frac{78}{500} P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 4 = \frac{F r e q u e n c y o f 4}{t o t a l t h r o w s} = \frac{150}{1000} = \frac{75}{500} P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 5 = \frac{F r e q u e n c y o f 5}{t o t a l t h r o w s} = \frac{176}{1000} = \frac{88}{500} P r o b a b i l i t y o f g e t t i n g o u t c o m e a s 6 = \frac{F r e q u e n c y o f 6}{t o t a l t h r o w s} = \frac{188}{1000} = \frac{94}{500}$
Regards,

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