Using the probability distribution of outcomes given from Problem 1.2, calculate

Question

Using the probability distribution of outcomes given from Problem 1.2, calculate the probability that two throws of a pair of dice have the outcomes (order of outcome does not matter): 8 + 12 9 + 11 10 + 10 From the pattern above, find the probability of outcome 20 from two throws of two dice. Given 3 distinguishable particles each of which can be in any of 4 states with different energies E_1, E_2, E_3, E_4, answer the following: What is the total number of ways to distribute the particles amongst the states? How many states of the system have total energy E_1 + E_2 + E_4? How does(a) and (b) change if the 3 particles are indistinguishable?

Explanation / Answer

When two dice are thrown, there are total 36 following cases :

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),...,(2,6),(3,1),...,(3,6),(4,1),(4,1),.....(4,6),(5,1),..........(5,6),(6,1),...(6,6)]

Now out of these 36 pairs, pairs (3,5),(5,3),(2,6),(6,2),(4,4) give sum 8 and pair (6,6) gives sum 12.

Now as probability of an event = total favorable case/total cases

so P(sum 8)= 5/36 and p(sum 12) = 1/36

Now P( sum 8 or 12 on two throw) = P( sum 8 or sum 12 in first throw) + P ( sum 12 or sum 8 in second throw)

=(5/36)+(1/6) +(5/36)+(1/36) = (5+5+1+1)/36= 12/36=1/3

This is the asnwer of part (a) of question (3).

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