Normal Approximation and Binomial Random Variable

Question

A fair pair of dice is rolled 120 times. Find the probability that the same point comes up on both dice at least 25 times. Pr[X ≥ 25]

Answer 1

roll 2 dice, 120 times

what's probability at some point "same" two outcomes come up on both dice at least 25 times?

does that mean same two 2's 25 times, or just tow of the same out come 25 times (as in two 2's, two 3's or two of 4's, 5's 6's or 1's??

try the 2nd interpretation, same two 2's or same other same outcomes 25 times

each time there are 6 ways out of 36 to get the same number on 2 dice P(same on one roll) = 6/36 = 1/6.

P(same on two rolls) = 1-(5/6)^2 = 1 - 25/36 = 11/36

P (same on 25 rolls) = 1- (5/6)^25 = about .9895 = 98.95%

P(same once on 120 rolls) = .9999999997 = virtually certain

120/25 = 4.8

1- (5/6)^4.8 = about .58% .58^4 = about 11%

= P(a double every 4.8 rolls of the dice, about every 5) that would be 25 doubles in 120 rolls, 120/4.8 = 25

or use the binomial as either you get a double or you don't just 2 alternatives,which are disjoint, not overlapping and exhausted the sample space

1/6 of the time you do, 5/6 you don't

np = 120 x 1/6 = 20 expect times you get a double on 120 rolls, but you want 25 or more

find the standard deviation = sd = sqrVar. = sqr(npq) = sqr(20x5/6) = sqr50/3 = about 4.08

z score = (25-20)/4.08 = about 1.23

find the area under the curve for z>1.23

use z tables

area = about .11 = 11%