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Consider the following scenarios, where we apply probability to a game of flipping coins. In the game, we flip one coin each round. The game will not stop until two consecutive heads appear. (a) What is the probability that the game ends by flipping exactly five coins? (b) Given that the game ends after flipping the fifth coin, what is the probability that three heads appear in the sequence?

Sagot :

sandlee09

Answer:

3.125%

0%

Step-by-step explanation:

A regular coin has two sides, meaning that each side has the same probability of landing which is 50%. Therefore, for the game to end exactly on the fifth flip it would need to land 3 times in a row on tails and the last 2 times on heads. Therefore, to get this probability we need to multiply 0.50 (50%) 5 times

0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.03125 or 3.125%

Now the probability of three heads appearing in a sequence is 0% because after the second head in a row the game will automatically end.

Part(a): The probability that the game ends by flipping exactly five coins is [tex]0.5[/tex]

Part(b):  The probability that three heads appear in the sequence is [tex]0.5[/tex].

Probability:

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. How likely they are to happen, using it.

Part(a):

The Probability of games ends by flipping exactly five coins is,

[tex]P((TTTHH),(HTTHH),(THTHH),(HHTHH))=(1-p)^3p^2+2(1-p)^2p^3+(1-p)p^4[/tex]

Where [tex]p=P(H)[/tex]

When the coin is fair so the probability is,

[tex]P(H)=\frac{1}{2}[/tex]

P(game ends by flipping exactly five coins)=[tex]\frac{2^2}{2^5} =0.125[/tex]

Part(b):

The Probability of games ends by flipping exactly three heads is,

[tex]\frac{2(1-p)^2p^3}{(1-p)^3p^2+2(1-p)^2p^3+(1-p)p^4}[/tex]

Where [tex]p=P(H)[/tex]

When [tex]p=0.5[/tex] then the above probability is [tex]0.5[/tex]

Learn more about the topic of Probability:

https://brainly.com/question/6649771

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