# A coin is tossed 5 times. Probability of getting at least 3 heads is $\dfrac{1}{{2}^{x}}$. What is the value of x?

Last updated date: 16th Mar 2023

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Hint: In order to solve such type of question firstly we have to find out the probability of exactly 3 heads, probability of getting exactly 4 heads and probability of getting all three heads then we will easily get the probability of getting at least three heads using the formula $^n{C_r}{a^r}{b^{n - r}}$

Complete step-by-step answer:

We know that,

Probability of getting a head when a coin is tossed $ = $ probability of getting a tail when a coin is tossed$a = b = \dfrac{1}{2}$.

We have given that,

A coin is tossed $5$ times. Therefore $n = 5$ and probability of getting at least $3$ heads $r = 3,4,5.$

Probability of getting at least three heads$ = $ Probability of exactly 3 heads $ + $ Probability of getting exactly 4 heads$ + $ Probability of getting all three heads.

Using formula,

$^n{C_r}{a^r}{b^{n - r}} - - - - - \left( 1 \right)$

Therefore,

$^5{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{5 - 3}}{ + ^5}{C_4}{\left( {\dfrac{1}{2}} \right)^4}{\left( {\dfrac{1}{2}} \right)^{5 - 4}}{ + ^5}{C_5}{\left( {\dfrac{1}{2}} \right)^5}{\left( {\dfrac{1}{2}} \right)^{5 - 5}}$

Or $^5{C_3}{\left( {\dfrac{1}{2}} \right)^5}{ + ^5}{C_4}{\left( {\dfrac{1}{2}} \right)^5}{ + ^5}{C_5}{\left( {\dfrac{1}{2}} \right)^5}$

Using formula $^n{C_r}{a^r}{b^{n - r}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$\left( {\dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} + \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}}} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\left( {\dfrac{{5 \times 4}}{{2 \times 1}} + \dfrac{5}{1} + 1} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\left( {10 + 5 + 1} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\dfrac{{16}}{{32}} = \dfrac{1}{2}$

Therefore, $x = 1$

Note: Whenever we face these types of questions the key concept is that simply we will understand the given part to calculate the value of $a,b,n,r$. Then we have to substitute this in this formula $^n{C_r}{a^r}{b^{n - r}}$ and we will get our desired answer.

Complete step-by-step answer:

We know that,

Probability of getting a head when a coin is tossed $ = $ probability of getting a tail when a coin is tossed$a = b = \dfrac{1}{2}$.

We have given that,

A coin is tossed $5$ times. Therefore $n = 5$ and probability of getting at least $3$ heads $r = 3,4,5.$

Probability of getting at least three heads$ = $ Probability of exactly 3 heads $ + $ Probability of getting exactly 4 heads$ + $ Probability of getting all three heads.

Using formula,

$^n{C_r}{a^r}{b^{n - r}} - - - - - \left( 1 \right)$

Therefore,

$^5{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{5 - 3}}{ + ^5}{C_4}{\left( {\dfrac{1}{2}} \right)^4}{\left( {\dfrac{1}{2}} \right)^{5 - 4}}{ + ^5}{C_5}{\left( {\dfrac{1}{2}} \right)^5}{\left( {\dfrac{1}{2}} \right)^{5 - 5}}$

Or $^5{C_3}{\left( {\dfrac{1}{2}} \right)^5}{ + ^5}{C_4}{\left( {\dfrac{1}{2}} \right)^5}{ + ^5}{C_5}{\left( {\dfrac{1}{2}} \right)^5}$

Using formula $^n{C_r}{a^r}{b^{n - r}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$\left( {\dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} + \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} + \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}}} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\left( {\dfrac{{5 \times 4}}{{2 \times 1}} + \dfrac{5}{1} + 1} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\left( {10 + 5 + 1} \right){\left( {\dfrac{1}{2}} \right)^5}$

Or $\dfrac{{16}}{{32}} = \dfrac{1}{2}$

Therefore, $x = 1$

Note: Whenever we face these types of questions the key concept is that simply we will understand the given part to calculate the value of $a,b,n,r$. Then we have to substitute this in this formula $^n{C_r}{a^r}{b^{n - r}}$ and we will get our desired answer.

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