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5 votes
A coin is loaded so that the probability of a head occurring on a single toss is 5/6. In five tosses of the​ coin, what is the probability of getting all heads or all​ tails? The probability of all heads or all tails is____ ​(Round to three decimal places as​ needed.)

asked
User Mehmetx
by
7.7k points

2 Answers

4 votes

Final answer:

The probability of getting all heads or all tails in five tosses of a loaded coin with a probability of 5/6 for a head is 0.401.

Step-by-step explanation:

To find the probability of getting all heads or all tails in five tosses of a loaded coin with a probability of 5/6 for a head, we need to calculate the probability of getting either all heads or all tails.

  • The probability of getting all heads in five tosses is (5/6)⁵.
  • Similarly, the probability of getting all tails in five tosses is also (1/6)⁵.
  • Since we want to find the probability of getting either all heads or all tails, we can add these probabilities together:
  • P (all heads or all tails) = (5/6)⁵ + (1/6)⁵ = 0.401
answered
User Palak Arora
by
8.0k points
3 votes

Final Answer:

The probability of getting all heads or all tails in five tosses of the loaded coin is
\(0.041\).

Step-by-step explanation:

For a loaded coin with a probability of landing heads in a single toss as \(5/6\) and \(1/6\) for tails, we're determining the probability of obtaining either all heads or all tails in five tosses.

The probability of getting all heads in five tosses is \((5/6)^5\) because the events are independent. Similarly, the chance of getting all tails in five tosses is \((1/6)^5\). The probability of getting all heads or all tails is the sum of these two individual probabilities.


\(P(\text{all heads}) = \left((5)/(6)\right)^5 = 0.401\) (approximately)\(P(\text{all tails}) = \left((1)/(6)\right)^5 = 0.000128\) (approximately)Therefore, the probability of getting all heads or all tails in five tosses is \(0.401 + 0.000128 = 0.041\) (approximately).

Understanding probability in simple events like coin tossing demonstrates the application of probability rules, especially with independent events. Such exercises help to comprehend the chances of specific outcomes occurring over multiple trials.

answered
User Dees Oomens
by
8.7k points
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