Three unbiased coins are tossed together find the probability of getting one head

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Solution

When 3 unbaised coins are tossed, the total number of events = 8 They are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} (1) Number of favourable events (getting at least two heads) = 4 Probability of getting at least two heads = 4/8 = 1/2 (2) At most two heads That is no head or one head or 2 heads Probability of getting at most two heads = 7/8 3) Two head = 3 (HHT,HTH,THH) (adsbygoogle = window.adsbygoogle || []).push({}); Probability of getting no head = 3/8.

The probability of getting exactly 2 heads is 12

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The probability of getting exactly two heads is 38

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The probability of getting at least 1 tail is 38

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The probability of getting exactly one tail is 38

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Solution

The correct option is DThe probability of getting exactly one tail is 38Total possible cases = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} (where H= Heads, T= Tails) Number of exactly 2 heads = {HHT, HTH,THH} = 3 ∴ P (exactly 2 heads) = 38Number of exactly 1 tail = {HHT, HTH, THH} = 3 ∴ P(exactly one tail) =38 (adsbygoogle = window.adsbygoogle || []).push({}); Number of cases atleast one tail = {HHT, HTH, HTT, THH, THT, TTH, TTT} = 7∴ P(at least one tail) = 78

Answer

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Hint: To solve this problem we need to first write the sample space of all the possible outcomes and then check the favorable outcomes that satisfy the given condition. After getting the favorable outcomes and total number of possible outcomes substitute them in the probability formula.
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\].

Complete step-by-step answer:
Let us first look at the basic definitions related to probability.
TRIAL: Let a random experiment be repeated under identical conditions then the experiment is called a trial.
OUTCOME: A possible result of a random experiment is called its outcome.
SAMPLE SPACE: The set of all possible outcomes of an experiment is called the sample space of the experiment and is denoted by S.
SAMPLE POINT: The outcome of an experiment is called sample point.
EVENT: A subset of the sample space associated with a random experiment is said to occur, if any one of the elementary events associated to it is an outcome.
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
Unbiased coin means that the probability of heads is the same as the probability of tails.
Let us assume that tossing of three coins as an event A.
Now, let us write the sample space when three coins are tossed together.
Let us denote the occurrence of heads when coin is tossed as H and the occurrence of tails when tossed as T.
Sample space(S) when three coins are tossed together is:
S = {H, H, H}, {H, H, T}, {H, T, T}, {T, H, H}, {T, T, H}, {T, H, T}, {H, T, H}, {T, T, T}.
The sample space (S) denotes the total number of possible outcomes.
Given, the condition is that all the coins show up heads which happens only once {H, H, H}.
So, number of favorable outcomes (m) = 1,
Total number of possible outcomes (n) = 8.
Now, by applying the probability formula we get,
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{m}{n}\]
\[\Rightarrow m=1,n=8\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{1}{8}\]
\[\therefore P\left( A \right)\text{=0}\text{.125}\].
So, the required probability is 0.125

Note: While writing the sample space one needs to make sure that you consider all the possible outcomes without fail because neglecting any one of the possible outcomes causes the entire result to be wrong.
For suppose if you forgot to consider the possibility of {H, T, H} and then when you calculate the total number of possible outcomes we get,
Total number of possible outcomes = 7
Number of favorable outcomes = 1
\[P\left( A \right)\text{=}\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]
\[\Rightarrow P\left( A \right)\text{=}\dfrac{m}{n}\]
\[\Rightarrow m=1,n=7\]
\[\therefore P\left( A \right)\text{=}\dfrac{1}{7}\].
Hence, the result obtained here is completely wrong.

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