Question

If a coin is tossed 3 times in succession,then find the probability of obtaining tail at least once​

Answer 1

Answer:

Sample Space : { HHH, HTH, HHT, THH, TTT, THT, TTH, HTT }

Event ( Getting tail at least once } = { HTH, HHT, THH, THT, TTH, HTT, TTT }

Since "Atleast" is used, it refers to one and above.

E.g. Atleast 5 means 5 and above.

Hence Event has 7 possibilities.

⇒ Probability = Possibilities / Sample Space = 7 / 8

Hence Probability of getting atleast 1 tail is 7/8.

Answer 2

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(at least 1 tail) = 1 - P(no tails) as probabilities add up to 1.

P (no tails) = P(head) & P(head) & P(head) / P(head, head, head)

P(head) = 1/2.

For two mutually exclusive (independent) events to happen consecutively, their probabilities are multiplied together (called the AND rule).

Therefore, P(head, head, head) = 1/2 x 1/2 x 1/2 = 1/8

$<b>P(at least 1 tail) = 1 - 1/8 = 7/8</b>$

This method is done assuming that the coin is fair and unbiased so the chances of getting headsand tails are the same.

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