five defective bulbs are accidently mixed with 20 good ones. it is not possible to just look at a bulb and tell whether or not it is defective. find the probability distribution from this lot.
given : five defective bulbs are accidently mixed with 20 good ones.to find : probability distribution from this lotformula used :five defective bulbs are accidently mixed with 20 good ones.total number of bulbs = 25x denote the number of defective bulbs drawnx can draw the value 0 , 1 , 2 , 3 , 4.since the number of bulbs drawn is 4, n = 4p(0) = p(getting a no defective bulb) = = = p(1) = p(getting 1 defective bulb and 3 good ones) = = p(1) = = p(2) = p(getting 2 defective bulbs and 2 good one) = p(2) = = = p(3) = p(getting 3 defective bulbs and 1 good one) = p(3) = = p(4) = p(getting all defective bulbs) = = = p(4) = the probability distribution table is as follows,
?answer :




step 1 of 6:


given, in a sample of 20 weld failures, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal.


let x follows binomial distribution with probability mass function is




p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


where,
n = sample size
 = 20
x = random variable
p = probability of success
 = 15% 
 = 0.15
q = 1 - p (probability of failure)


mean of the binomial distribution is 


 = np


variance of the binomial distribution is 


 = npq


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step 2 of 6:




1. the claim is to find the probability that exactly five of them are base metal failures


we have to find p(x=5) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 5


p(x = 5) = (1 - 0.15


 = (15504) (0.0000759) (0.087354)


 = 0.1028


hence, the probability that exactly five of them are base metal failures is 0.1028.


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step 3 of 6:




1. the claim is to find the probability that fewer than four of them are base metal failures


we have to find p(x < 4) = p(x = 0) + p( x = 1) + p(x = 2) + p(x = 3) 






p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0, 1, 2 and 3


p(x < 4) = (1 - 0.15+ (1 - 0.15+


 (1 - 0.15+ (1 - 0.15


 = (1) (1) (0.03876) + (20) (0.15) (0.0456) +
 (190) (0.0225) (0.053646) + (1140) ( 0.003375) (0.0631134)


 = 0.03876 + 0.1368 + 0.2293 + 0.2428


 = 0.6477


hence, the probability that fewer than four of them are base metal failures 0.6477


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step 4 of 6:




c) the claim is to find the probability that none of them are base metal failures.


we have to find p(x=0) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0


p(x = 0) = (1 - 0.15


 = (1) (1) (0.03876) 


 = 0.03876


hence, the probability that none of them are base metal failures is 0.03876


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step 5 of 6:




d) the claim is to find the mean number of base metal failures..


 mean of the binomial distribution is = np


where, n = 20, and p = 0.15


substitute these values in 


then, = (20) (0.15)


 = 3


hence, the mean number of of base metal failures is 3.


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step 6 of 6:




e) the claim is to find the standard deviation of the number of base metal failures..


standard deviation of the binomial distribution is 


where, n = 20, p = 0.15 and q = 1 - p


 = 1 - 0.15


 = 0.85


substitute these values in 


then, 


 


 = 1.596


hence, the standard deviation of the number of base metal failures is 1.596.


\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ total number of cards in a well-shuffled pack =52total number of queens and kings cards =4+4=8therefore, there are 44 cards (52−8) that are neither king nor queen.total number of favorable outcomes  =44  
probability is given by,p(a)=no. of total outcomesno. of possible outcomes​  
required probability =5244​=1311​ can't be determined can you list the things that go wrong and how they should look? Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3 it could be a random drop or a glitch. if you've got the item and nothing else is wrong with the game, enjoy! because i only had 10 dollars on me, my choices between having a burger meal and a pizza meal for lunch were mutually exclusive.

this light cannot be both on and off at the same time because they are mutually exclusive events.

you can, of course, have your cake and eat it too. those two thing are not mutually exclusive. giving a name to checkboxes means grouping them together to the same question

if you have 10 questions, and each has 5 checkboxes, and you don’t group them together, that would mean having 50 checkboxes lumped together - maybe you have also pretty similar values etc…

you give each checkboxes group… maybe in you limited experience with t1dm you know of one or more with this appearance. in the spectrum of those affected i find you cannot generalize this description to them at all. you would not be able to pick most out of a crowd. see more... let's be clear, here: luminosity, "lumen output", "light output", or whatever you want to call the light intensity of a bulb is not a direct function of filament temperature. that's a gross ignorance and nothing more.
incandescents are rated by the power they use and nothing else. the color temperature is a function of the filament and the power applied. most filaments are tungsten so the filament can never heat above 3,700 kelvin, the melting point of tungsten.

so, most filaments are limited to 2,000 to 3,300 kelvin. this is the resulting color temperature and why some lights are whiter and some are more yellow., the latter tending to last longer since they are cooler or "warm" light.
in other words, a 100 watt bulb and a 40 watt bulb operate at the same filament temperature if they are both the same type of bulb, but the 100 watt bulb produces more of it. the "light output" , is not a function of filament temperature as veeyesvee and tony rb asserted.
so, a 60 watt bulb, tony, better produce more lumens than a 25 watt bulb, or what's the point of the other 35 watts? can someone design a really ineffecient transducer? sure... a toaster makes a terrible light source considering the power it consumes.
the real answer lies in the fact that the filament is not a linear resistance ( see below).

the hotter the bulb gets, the more efficient it becomes in producing light and a 60 watt bulb is about 14.2 lumens per watt where a 25 watt bulb is 8 lumens/watt.

the amount of light produced by a bulb depends on the ratings vs applied. since all you can vary is the voltage:
raise the voltage ratio to the 3.5 power to get equivalent luminosity.
life would be the -12th power of the voltage ratio.
and current is the .55 power of the voltage ratio.

(v2/v1)^3.5 = .235 (since 200lm/850lm = .235) 
(v2/v1)^3.5 = .235 ==> v2 = 120 * .661 = 79.34 volts

(v2/v1)^.55 = i2/.5 (60/120 = .5) ==> i2 = .398 a

79.34 x .398 = 31.6 w
or v2/v1 * (v2/v1) ^.55 = .52 power ratio and 60 * .52 = 31.6 w

so, the 60 watt bulb would need 32 watts of power to generate 200 lumen.