six dice are thrown 729 times. how many times do you expect atleast 3 dice to show a five or a six?

**1** Let the probability of occurrence of the event exactly in 3 dice be referred to as P (E, 3). **2** Let the event be, getting 4 or 5 when a die is thrown.**3** Probability of occurrence of the event in a single throw is **4** **5** Probability of occurrence of the event in at least 3 dice = P (E, 3) + P (E, 4) + P (E, 5) + P (E, 6)**6** This is a binomial situation and hence using the formula P (E, ^{C}^{r} · ^{n - r}**7** The required probability = (6^{C}3)(^{3} (^{3} + (6^{C}4)(^{4}(^{2} + (6^{C}5)(^{5}(^{C}6)(^{6} **8** Required probability = (

[Simplify.]**9** Out of 729 throws, the number of times atleast 3 dice will show a 5 or 6 is 729 × **10** So, we can expect at least 3 dice to show a 5 or 6 for 233 times