GATE CSE 2016 Set 2 | Question: 05

Question

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 _________.

Answer 1

Given that the 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 $0.4$. i.e.,

$Prob(100+ \mid Type1) = 0.7$

$Prob(100+\mid Type 2) = 0.4$

$Prob(100+) = Prob(100+ \mid Type1) \times Prob(Type1) + Prob(100+ \mid Type2) \times Prob(Type2)$

$\quad \quad = 0.7 \times 0.5 + 0.4 \times 0.5 = 0.55.$

Comments

tree approach is best fr this kind of question....

---

Can you please once draw the tree for it?

---

“This question is simply the application of Total probability law” .

Answer 2

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

Comments

nice approach :)

Answer 3

According to Total Probability theorem:-

 

$P(>100) = P(T_1)\times P(\frac{> 100}{T_1})+ P(T_2)\times P(\frac{> 100}{T_2})$

$P(>100) = 1/2\times 0.7+1/2\times 0.4 = 0.55$

Answer 4

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

Comments

Tree diagram and this problem is solved in an instant.....