p fail = .2
p pass = .8
p five pass = .8^5 = .32768
HOW CAN YOU USE DISCRETE PROBABILITY TO CALCULATE THE PROBABILITY OF TESTING 5 DEVICES SUCH THAT ALL 5 END UP BEING NONDEFECTIVE, GIVEN THERE ARE 20 DEFECTIVE DEVICES TOTAL OF THE ORIGINAL 100 DEVICES? WHAT IS THE SAMPLE SPACE?
2 answers
The finite population has 100 devices, out of which 80 are non-defective.
At each draw, both the total number devices and the number of non-defective device decrease, we have the running probabilities as:
{80/100, 79/99, 78/98, 77/97, 76/96}
the product of which give 0.31931, very close the the earlier estimate of 0.32.
Alternatively, the hypergeometric distribution may be used, which gives
P(5 nondefectives)
=C(20,0)*C(80,5)/C(100,5)
=0.31931 as before.
To find Ω, we determine that there 2 possible outcomes for each draw, hence for 5 draws, the sample space is 2^5=32.
At each draw, both the total number devices and the number of non-defective device decrease, we have the running probabilities as:
{80/100, 79/99, 78/98, 77/97, 76/96}
the product of which give 0.31931, very close the the earlier estimate of 0.32.
Alternatively, the hypergeometric distribution may be used, which gives
P(5 nondefectives)
=C(20,0)*C(80,5)/C(100,5)
=0.31931 as before.
To find Ω, we determine that there 2 possible outcomes for each draw, hence for 5 draws, the sample space is 2^5=32.
1 answer