Question

Six bags of cement, each marked 10 kg, actually contained the following weights of cement (in kg): ., ., ., ., ., . Find the probability that any of these bags chosen at random contains less than 10 kg of cement.​

Answer 1

Given : Six bags of cement, each marked 10 kg, actually contained the following weights of cement (in kg): 10.1, 9.95, 10.4, 9.93, 10.8, 10.0

To Find  : the probability that any of these bags chosen at random contains less than 10 kg of cement.

1/2

1/3

2/3

1/6

Solution :

S = {10.1, 9.95, 10.4, 9.93, 10.8, 10.0 }

number of bags = 6

n(S) = 6

E =  Bag contains less than 10 kg of cement.

E = {9.95 , 9.93 }

n(E) = 2

probability that any of these bags chosen at random contains less than 10 kg of cement.  = n(E)/n(S)

= 2/6

= 1/3

1/3 is  the probability that any of these bags chosen at random contains less than 10 kg of cement.

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Answer 2

${\underline{ \mathtt{\red{A} \green{n}\mathtt\blue{s} \purple{w} \mathtt \orange{e}\pink{r}}}}$

Total number of bags = 6

Bag whose weight is less than 10 kg = 9.95, 9.93

So number of bags less than 10 kg = 2

So required probability is 2/6 = 1/3

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