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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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Given, Shop has equal number of bulbs 

$\therefore$ $P(Type 1)$ = $P(Type 2)$ = $0.5$

Probability of bulb lasting more than 100 hr given that it’s $Type – 1$ = $P$($>100hr$ | $Type 1$) = $0.7$

Probability of bulb lasting more than 100 hr given that it’s $Type – 2$ = $P$($>100hr$ | $Type 2$) = $0.4$

$\therefore$ Probability that an LED bulb lasts more than 100 hrs.:

$P(> 100 hr)$ = $P$($>100hr$ | $Type 1$) $\times$ $P(Type 1)$ $+$  $P$($>100hr$ | $Type 2$) $\times$ $P(Type 2)$

$P(> 100 hr)$ = $0.7 \times 0.5 + 0.4 \times 0.5 = 0.35 + 0.20$ = $0.55$ [Ans]

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1 votes


The bulbs of Type 1, Type 2 are same in number.
So, the probability to choose a type is 1/2.
The probability to choose quadrant ‘A’ in diagram is
P(last more than 100 hours/ type1) = 1/2 × 0.7
P(last more than 100 hours/ type2) = 1/2 × 0.4
Total probability = 1/2 × 0.7 + 1/2 × 0.4 = 0.55

Answer:0.55

Answer:

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