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# A die is rolled 12 times. Find the probability of rolling no more than 4 fives.

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A die is rolled 12 times. Find the probability of rolling no more than 4 fives.

Feb 21, 2015

#2
+95360
+10

P(no more thatn 4 fours) = P(no fours)+P(1four)+P(2 fours)+P(3 fours)+P(4 fours)

$$\\\left(\frac{5}{6}\right)^{12} \;\;+\;\;^{12}C_1\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{11} \;\;+\;\;^{12}C_2 \left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{10}\\\\ \;\;+\;\;^{12}C_3\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{9} \;\;+\;\;^{12}C_4\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{8}$$

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Feb 21, 2015

#1
+94558
+10

This means we either rolll no fives, one five, two fives, three fives or 4 fives

The probability of rolling no fives is C(12,0)(1/6)^0*(5/6)^12 = 0.1121566547846151

The probability of rolling one five is C(12,1)(1/6)^1*(5/6)^11 = 0.0000000044516784

The probability of rolling two fives is C(12,2)(1/6)^2*(5/6)^10 = 0.2960935686313838

The probability of rolling three fives is C(12,3)(1/6)^3*(5/6)^9 = 0.1973957124209225

The probability of rolling four fives is C(12,4)(1/6)^4*(5/6)^8 = 0.0888280705894151

And summing these we have......about 69.4%  chance of rolling no more than 4 fives

Feb 21, 2015
#2
+95360
+10

P(no more thatn 4 fours) = P(no fours)+P(1four)+P(2 fours)+P(3 fours)+P(4 fours)

$$\\\left(\frac{5}{6}\right)^{12} \;\;+\;\;^{12}C_1\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{11} \;\;+\;\;^{12}C_2 \left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{10}\\\\ \;\;+\;\;^{12}C_3\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{9} \;\;+\;\;^{12}C_4\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{8}$$

Melody Feb 21, 2015
#3
+94558
+5

I beat you, Melody........!!!!!

Feb 21, 2015
#4
+95360
+5

Yes, you beat me fair and square    LOL   :)))

AND I DON'T CARE - TAKE THAT !!!

Feb 21, 2015
#5
+94558
+5

LMAO......!!!!

Feb 21, 2015
#6
0

what would be the answer to no more than 4 fives in 10 rolls

Jul 11, 2016