Three coins are tossed in the air and two of the coins land with heads face upwards. What are the chances on the next toss of the coins that at least two of the coins will land with heads face upwards again?

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Three coins are tossed in the air and two of the coins land with heads face upwards. What are the chances on the next toss of the coins that at least two of the coins will land with heads face upwards again?

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Assuming the previous coin tosses have no influence on the outcome of any future coin tosses the probability of having atleast 2 of the 3 coins facing upwards is:

[Events favoring the given event] / [Total number of possible events]

Total possible events = 2*2*2 = 8.

number favorable outcomes = 3!/2! + 3!/3! = 6/2 + 6/6 = 3(2 heads and 1 tail) + 1(All heads outcome) = 4.

ie. Probability = 4/8 = **0.5** or **50%**.

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