Suppose you roll a fair six-sided die 25 times. What is the probability that you roll 5 or more 6’s on that die? A. 0.3883

B. 0.5937

C. 0.5

D. 0.4063

D. P(5+ 6's) = 0.4063

Step-by-step explanation:

Binomial distribution.

For the distribution to be applicable, the experiment must

1.Have a know and constant number of trials

2.Probability of success of each trial remains constant (and known if available)

3.Each trial is a Bernoulli trial, i.e. with only two outcomes, success or failure.

4.Independence between trials.

Let  

n = number of trials  = 25

p = probability of success of each trial  = 1 / 6

x = number of successes  (0 ≤ x ≤ n)  = 5

C(n,x) = number of combinations of picking x identical objects out of n

Applying binomial distribution

P(x,n) = probability of x successes in an experiment of n trials.

= C(n,x) * p^x * (1-p)^(n-x)

For n = 25 trials with probability of success (roll a 6) = 1/6

and x = 5,6,7,8,...25

It is easier to calculate the complement by

P(5+ 6's) = 1 - P(<5 6's)

= 1 - ( P(0,25) + P(1,25) + P(2,25) + P(3,25) + P(4,25) )

1- (

    P(0,25) = C(25,0) * (1/6)^0 * (5/6)^25 = 0.0104825960103961

    P(1,25) = C(25,1) * (1/6)^1 * (5/6)^24 = 0.05241298005198051

     P(2,25) = C(25,2) * (1/6)^2 * (5/6)^23 = 0.1257911521247532

    P(3,25) = C(25,3) * (1/6)^3 * (5/6)^22 = 0.1928797665912883

    P(4,25) = C(25,4) * (1/6)^4 * (5/6)^21 = 0.2121677432504171

)

= 1 - 0.59373

= 0.40626

= 0.4063 (to 4th decimal place)

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: