Question 816315
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Hi,
Re Reply: note highlighted
Note: The probability of x successes in n trials is: 
P = nCx* {{{p^x*q^(n-x)}}} where p and q are the probabilities of success and failure respectively. 
In this case p = 1/6. q = 5/6 and n = 4  nCx = {{{n!/(x!(n-x)!)}}}

A die is thrown four times. Determine the probability of throwing the following.
 (a) zero 3's (Round your answer to four decimal places.) {{{(5/6)^4}}} = .4823
  P(x=0) = 4C0 = (5/6)^4
 (b) one 3 (Round your answer to four decimal places.) {{{highlight(4)(1/6)^1(5/6)^3}}}= .3858
  P(x = 1) = 4C1((1/6)^1(5/6)^3  
 (c) two 3's (Round your answer to four decimal places.) {{{6(1/6)^2(5/6)^2}}}= .1157
 P(x = 2) = 4C2((1/6)^2(5/6)^2 
 (d) three 3's (Round your answer to four decimal places.){{{4(1/6)^3(5/6)^1}}} = .0154
 P(x = 3) = 4C3((1/6)^3(5/6)^1
 (e) four 3's (Round your answer to five decimal places.){{{(1/6)^4(5/6)^0}}}= .00077
 P(x = 4) = 4C4((1/6)^4(5/6)^0
Note: The probability of x successes in n trials is: 
P = nCx* {{{p^x*q^(n-x)}}} where p and q are the probabilities of success and failure respectively. In this case p = 1/6. q = 5/6 and n = 4  nCx = {{{n!/(x!(n-x)!)}}}