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Question: A coin is tossed 5 times. Probability of getting at least 3 heads is \(\dfrac{1}{{2}^{x}}\). What is...

A coin is tossed 5 times. Probability of getting at least 3 heads is 12x\dfrac{1}{{2}^{x}}. What is the value of x?

Explanation

Solution

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 nCrarbn−r^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 tosseda=b=12a = b = \dfrac{1}{2}.
We have given that,
A coin is tossed 55 times. Therefore n=5n = 5 and probability of getting at least 33 heads r=3,4,5.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,
nCrarbn−r−−−−−(1)^n{C_r}{a^r}{b^{n - r}} - - - - - \left( 1 \right)
Therefore,
5C3(12)3(12)5−3+5C4(12)4(12)5−4+5C5(12)5(12)5−5^5{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 5C3(12)5+5C4(12)5+5C5(12)5^5{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 nCrarbn−r=n!r!(n−r)!^n{C_r}{a^r}{b^{n - r}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
(5!3!(5−3)!+5!4!(5−4)!+5!5!(5−5)!)(12)5\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 (5×42×1+51+1)(12)5\left( {\dfrac{{5 \times 4}}{{2 \times 1}} + \dfrac{5}{1} + 1} \right){\left( {\dfrac{1}{2}} \right)^5}
Or (10+5+1)(12)5\left( {10 + 5 + 1} \right){\left( {\dfrac{1}{2}} \right)^5}
Or 1632=12\dfrac{{16}}{{32}} = \dfrac{1}{2}
Therefore, x=1x = 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,ra,b,n,r. Then we have to substitute this in this formula nCrarbn−r^n{C_r}{a^r}{b^{n - r}} and we will get our desired answer.