Question

Cody buys a soda that offers another soda free if he is lucky. The cap reads '1 in 6 wins!', meaning that each soda has a 1/6 probability of winning. Cody sees this and buys six of these sodas, thinking he is guaranteed a seventh. What is the true probability he will win at least one more soda? Express your answer as decimal to the nearest hundredth.

Answer 1

Answer: 0.26

Step-by-step explanation:

Binomial probability formula :-

$P(X=x)=^nC_x\ p^x\ (1-p)^{n-x}$, where P(x) is the probability of getting success in x trials , n is total number of trials and p is the probability of getting success in each trial.

Given : The probability of winning = $\dfrac{1}{6}$

Let X be the random variable that represents the number of sodas.

Since he is guaranteed that he will win one soda .

If Cody buys 6 sodas, then the probability that he will win at least one more soda will be :

$P(x\geq2 )=1-(P(0)+P(1))\\\\=1-(^6C_0\ (\dfrac{1}{6})^0\ (1-\dfrac{1}{6})^{6-0}+^6C_1\ (\dfrac{1}{6})^1\ (1-\dfrac{1}{6})^{6-1})\\\\=1-((\dfrac{5}{6})^6+(\dfrac{5}{6})^5)\approx0.26$

Hence, the true probability he will win at least one more soda =0.26