If five Fair coins are tossed find the probability that no more than two or tails

Question

Don Sumner · Accepted Answer

The problem is asking for the probability of getting no more than two tails when five fair coins are tossed.
Let's calculate the probability for each possible outcome:
- Number of Tails: 0
- Number of Heads: 5
- Number of ways this can occur:

(\binom{5}{0}) = 1

(choosing 0 tails out of 5 coins)
- Probability:

{(\frac{1}{2})}^{5} = \frac{1}{32}

- Number of Tails: 1
- Number of Heads: 4
- Number of ways this can occur:

(\binom{5}{1}) = 5

(choosing 1 tail out of 5 coins)
- Probability:

{(\frac{1}{2})}^{5} = \frac{1}{32}

- Number of Tails: 2
- Number of Heads: 3
- Number of ways this can occur:

(\binom{5}{2}) = 10

(choosing 2 tails out of 5 coins)
- Probability:

{(\frac{1}{2})}^{5} = \frac{1}{32}

To find the probability of no more than two tails, we need to sum up the probabilities of the above three cases:

P (no more than two tails) = P (0 tails) + P (1 tail) + P (2 tails)

P (no more than two tails) = \frac{1}{32} + \frac{1}{32} + \frac{1}{32} = \frac{3}{32}

Therefore, the probability of getting no more than two tails when five fair coins are tossed is

\frac{3}{32}

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If five Fair coins are tossed find the

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