A fair die is cast 4 times. Calculate the probability of obtaining at least two 6’s. Round to the nearest tenth of a percent

Question

Binomial distribution- probability and statistics

Answer 1

There are 1296 possible outcomes  [1111, 1112, 1113, 1114, ....6664, 6665, 6666].

 

Of these, 171 have 2 or more 6's.

 

The probability is 171/1296 =  0.13194

 

Rounded to the nearest tenth of a percent, that is P=13.2%.

 

 

Probability of getting a 6 when a dice is thrown once = 1/6.

 

Probability of getting 6 k times when thrown n times  =  (n C k)*(p^k)*(1-p)^n-k

 

    Here,  n=4 and k=2,3, 4

 

         (4 C 2)*(1/6)²(5/6)²    +    (4 C 3)*(1/6)³(5/6)¹  +  (4 C 4)*(1/6)⁴

                0.115740741        +   0.015432099       +       0.000771605

                                  0.131944444

Answer 2

p = .167 Since the probability of rolling a 6 on each roll is 1/6

X = 2 or more

n = 4 trials

 

So the probability of at least two sixes is P(X≥2)

 

You can use a calculator to find the cumulative probability or you can add P(x=2) + P(x=3) + P(x=4)