
A coin is tossed repeatedly until it shows its head. Let X be the number of tosses required to get head. Write the probability distribution of X.
Answer
615.6k+ views
Hint: In this problem, we will use the concept of conditional probability to write the probability distribution of X. If A and B are two events in a sample space S, then the conditional probability of A, with B given is defined as:
$P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$
Complete step-by-step answer:
Conditional probability is a measure of the probability of an event occurring given that another event has (by assumption, presumption, assertion or evidence) occurred. If the event of interest is A and the event B is known or assumed to have occurred, “the conditional probability of A given B”, or “ the probability of A under the condition B”, is usually written as $P\left( {}^{A}/{}_{B} \right)$.
This may be visualized as restricting the sample space to situations in which B occurs. The logic behind this equation is that if possible outcomes for A and b are restricted to those in which B occurs, this set serves as the new sample space.
When A and B are independent events, then:
\[P\left( {}^{A}/{}_{B} \right)=P\left( A \right)\,\,\,\,and\,P\left( {}^{B}/{}_{A} \right)=P\left( B \right)............(1)\]
When we flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of $\dfrac{1}{2}$.
In this question, it is given that a head should be obtained.
If we flip the coin once, then the probability of getting head is $\dfrac{1}{2}$.
But, if we don’t get head in one flip, then the probability of getting head in the second flip given as P(E) is:
$\begin{align}
& P\left( E \right)=P\left( A \right).P\left( {}^{B}/{}_{A} \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,=P\left( A \right).P\left( B \right) \\
\end{align}$ ( from equation 1 )
Where P(A) and P(B) are the probability of getting tails and heads respectively.
So, the probability distribution of X is given as:
Note: Students should note that here A and B both are independent events. The occurrence of one event does not depend on the other. We can use equation (1) while solving the problem only when A and B are independent events.
$P\left( {}^{A}/{}_{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$
Complete step-by-step answer:
Conditional probability is a measure of the probability of an event occurring given that another event has (by assumption, presumption, assertion or evidence) occurred. If the event of interest is A and the event B is known or assumed to have occurred, “the conditional probability of A given B”, or “ the probability of A under the condition B”, is usually written as $P\left( {}^{A}/{}_{B} \right)$.
This may be visualized as restricting the sample space to situations in which B occurs. The logic behind this equation is that if possible outcomes for A and b are restricted to those in which B occurs, this set serves as the new sample space.
When A and B are independent events, then:
\[P\left( {}^{A}/{}_{B} \right)=P\left( A \right)\,\,\,\,and\,P\left( {}^{B}/{}_{A} \right)=P\left( B \right)............(1)\]
When we flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of $\dfrac{1}{2}$.
In this question, it is given that a head should be obtained.
If we flip the coin once, then the probability of getting head is $\dfrac{1}{2}$.
But, if we don’t get head in one flip, then the probability of getting head in the second flip given as P(E) is:
$\begin{align}
& P\left( E \right)=P\left( A \right).P\left( {}^{B}/{}_{A} \right) \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,=P\left( A \right).P\left( B \right) \\
\end{align}$ ( from equation 1 )
Where P(A) and P(B) are the probability of getting tails and heads respectively.
So, the probability distribution of X is given as:
Note: Students should note that here A and B both are independent events. The occurrence of one event does not depend on the other. We can use equation (1) while solving the problem only when A and B are independent events.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

A moving boat is observed from the top of a 150 m high class 10 maths CBSE

