Please Help (100 points)!!
2 answers:
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Answer:
A) He should take the route with 4 crossings
B) The probability that he took the 4 crossing route is 0.2158
Step-by-step explanation:
Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.
Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1
The probability that the Professor is on time on route 1 is equal to
P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477
On the other hand, the probability that the professor is on time on route 2 is
P(X|B = 0) = 0.9² = 0.81
Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.
B) Lets call L the event 'The professor is late'. We know that
P(L|A) = 1-0.9477 = 0.0524
P(L|B) = 1-0.81 = 0.19
Also
P(A) = P(B) = 1/2 (this only depends on the result of the coin.
For the Bayes theorem we know, therefore that
Hence, the probability that he took the 4 crossing route is 0.2158.
We are given
total number of particles =4
and
total sum of charge=-36
Since, all the particles have the same charge
and total charge is negative
so, charge on each particle must be negative
now, we can use formula to find change on each particle
charge on each particle = (total sum of charge)/( total number of particles)
now, we can plug values
and we get
The charge on each particle is
units..............Answer