student2014
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- Oct 6, 2014
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Could you please check if my solution is correct?
If we have 1000 light led in a line, each with a probability of 30% to fail, what is the probability that 2 or more failed light led are next to each other?
- So since we have 1000 light led , We can have 300 failed light led ( 1000 * 0.3 = 300 ) .
- The distribution of these failed light led in the line can be 1 single light led , 2 light led next each other , 3 light led , …., ( can be up to 300 light led next to each other in which the probability is low )
- How can I calculate the probability that 2 or more light led are next to each other?
I tried calculating them separately using combination (Cn,r ):
Case 2 : P1 = C300,2/C1000,2
Case 3 : P2= C300,3/C1000,3
Up to case 30: P30= C300,30/C1000,30
then adding the probabilities of all cases is that correct ?
I do not have much knowledge in probability field so if there is other way to solve it I would appreciate if you mention it to me
Thank you
If we have 1000 light led in a line, each with a probability of 30% to fail, what is the probability that 2 or more failed light led are next to each other?
- So since we have 1000 light led , We can have 300 failed light led ( 1000 * 0.3 = 300 ) .
- The distribution of these failed light led in the line can be 1 single light led , 2 light led next each other , 3 light led , …., ( can be up to 300 light led next to each other in which the probability is low )
- How can I calculate the probability that 2 or more light led are next to each other?
I tried calculating them separately using combination (Cn,r ):
Case 2 : P1 = C300,2/C1000,2
Case 3 : P2= C300,3/C1000,3
Up to case 30: P30= C300,30/C1000,30
then adding the probabilities of all cases is that correct ?
I do not have much knowledge in probability field so if there is other way to solve it I would appreciate if you mention it to me
Thank you