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Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than $100$ hours given that it is of Type $1$ is $0.7$, and given that it is of Type $2$ is $0.4$. The probability that an LED bulb chosen uniformly at random lasts more than $100$ hours is _________.
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Suppose that a shop has an equal number of LED bulbs of two different types.  ==> Therefore

Probability of Taking Type 1 Bulb => 0.5

Probability of Taking Type 2 Bulb => 0.5

The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4.

Prob(100+ | Type1) => 0.7

Prob(100+| Type2) => 0.4

Prob(100+) => Prob(100+ | Type1) * Prob(Type1) + Prob(100+ | Type2) * Prob(Type2)

= 0.7 * .5 + .4 * .5

= 0.55

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I submitted 0.5 is it correct then? :/
Nope
Ok :/
tree approach is best fr this kind of question....
Can you please once draw the tree for it?
SIMPLY-

TYPE-1 bulbs ===>let total =10 then 7 are going for 100 years

TYPE-2 bulbs====>let total=10 then 4 are going for 100 years

TOTAL==>20 out of which 11 are going for 100 years ..THEREFORE probability=11/20=.55
nice approach :)

by using rule of total probility

p(>100 hrs.) = p(type 1)*p(>100hrs/type1) + p(type 2)*p(>100hrs/type2)

=(1/2)*(0.7) + (1/2)*(0.4) =0.55

According to Baye's theorem :-

P(E) = 1.1/2 = 0.55