Three people toss a fair coin and the odd one pays for coffee. If the coins all turn up the same, they are tossed again. Find the probability that fewer than 3 tossed are needed.

P(X < 3) = 0.9375

Step-by-step explanation:

If the coins tossed by the 3 people all turn up the same, it means that we will have either; HHH or TTT

H is head and T is tail.

Thus; the probability of failure in each trial is;

q = P(HHH) + P(TTT)

P(H) = ½ and P(T) = ½.

Hence;

q = P(HHH) + P(TTT) = (½ × ½ × ½) + (½ × ½ × ½) = ⅛ + ⅛ = ¼ = 0.25

In binomial probability, p = 1 - q

Thus; p = 1 - 0.25 = 0.75

Now to find the probability that fewer than 3 tosses are needed is given by the expression;

P(X < 3) = P(1) + P(2)

Now, because the toss trials are independent, it means that x = 1.

Since x = 1,we can make use of geometric distribution formula which is given by;

P(X = x) = p^(x) • q^(n - x)

Thus;

P(1) = 0.75^(1) × 0.25^(1 - 1)

P(1) = 0.75 × 1

P(1) = 0.75

P(2) = 0.75^(1) × 0.25^(2 - 1)

P(2) = 0.75 × 0.25

P(2) = 0. 75 × 0.25

P(2) = 0.1875

Thus;

P(X < 3) = 0.75 + 0.1875

P(X < 3) = 0.9375

idk

step-by-step explanation:

i don't know