Question

find the probability of throwing at most 2 sixes in 6 throws of a single die.


plzzzzzzzzz URGENT!!

Answer 1

Let event 

A : Getting a six 
B : Not getting a six 

P(A) = 1/6 

P (B) = 5/6

at most 2 sixes means to get 0,1 or 2 sixes 

probability of 0 six = ⁶C₀ (5/6)⁶ = (5/6)⁶
 
probability of 1 six = ⁶C₁ (5/6)⁵(1/6) = (5/6)⁵

probability of 2 six = ⁶C₂ (5/6)⁴(1/6)² = 1/2 (5/6)⁵

hence required probability = probability of 0 six + probability of 1 +                                                                                              six probability of 2 six

= (5/6)⁶ + (5/6)⁵ + 1/2 (5/6)⁵

= 7/3(5/6)⁵ ANSWER

Answer 2

Answer:(5/6)^4(35/18)