Ana, Betty and Carol are doing a competition. Ana can throw a die six times and aims to get at least one ‘6’. Betty can throw twelve times but needs to get at least two 6s. Carol has eighteen chances to get at least 3. Who has a better chance of winning the competition?
Let’s calculate the chances of each using binomial trials.
the probability of s successes in n rounds is nCs x ps x q(n-s)
Ana’s chances
The probability of getting at least one 6 in 6 tries is (1 – the probability of getting no success)
1 – 6C0 x (1/6) 0 x (5/6)(6-0) = 0.665
Betty’s chances
The probability of getting at least two 6s in 12 tries is (1 – the probability of getting no success – the probability of getting one success).
1 – 12C0 x (1/6) 0 x (5/6)(12-0) – 12C1 x (1/6) 1 x (5/6)(12-1) = 0.619
Carol’s chances
The probability of getting at least three 6s in 18 tries is (1 – the probability of getting no success – the probability of getting one success – the probability of getting two success).
1 – 18C0 x (1/6) 0 x (5/6)(18-0) – 18C1 x (1/6) 1 x (5/6)(18-1) – 18C2 x (1/6) 2 x (5/6)(18-2) = 0.597
