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.
Answers
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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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