You roll two dice 100 times and count the number of the sums...

You roll two dice 100 times and count the number of the sums of 6 obtained (note: P(6) = 5/36 = .1389) a) Find the probability you would get more than 10 sums of 6? Answer: b) Find the probability you get less than 6 sums of 6 in 100 rolls? c) On average how many sums of 6 should you get in 100 rolls of the dice?

Transcript

00:02 So we're given that a dice is thrown 100 times and the probability of getting a six is 5 over 36, right? so now they're going to ask us about the probabilities of getting how many sixes in those in a hundred number of throws, right? so in the first one, they ask us to calculate the probability that we get more than 10 sixes, right? so it means that we want the probability that x is greater than 10, right? and i've already established that what i'm working with is a binomial distribution, right, where n is 100.

00:38 And p, the probability of success, which is the probability of getting a six is five over 36, right? now, because our n is large, our n is large because it's greater than 30.

00:49 It means that i can convert this to be a normal, right, a normal distribution, such that i have x follows a normal right with mu is n p so np is 100 times 5 over 36 and by doing that i get um 13 .8 889 and the variance is n pq right so it's a hundred times five over 36 times one minus five for 36 and i get 11 .9 999 right right? so i'm just going to call this mu from now on in sigma.

01:43 It's a good idea to store this.

01:46 This is sigma squared to store these values, right, in your calculator so that you can just keep on referencing them.

01:54 Right.

01:54 So to get the probability that x is greater than 10, since we're dealing, since i'm not dealing with the normal distribution, i'm going to say that this is equal to 1 minus the probability that x is less than or equal to 10, right? and this is equal to one minus the probability that z is less than or equal to.

02:15 I'm now standardizing this thing.

02:17 So it's 10 minus mu, right, over the square root of my variance sigma squared, right? and when i evaluate this, i get one minus the probability that z is less than or equal to minus 1 .1 .1 .2...

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