Answer

Verified

448.8k+ views

Hint: In this question it is given that A’s skill is to B’s as 3 to 2, this means that A’s chances of winning a game is $\dfrac{3}{5}$ using the basic concept of probability that ${\text{Probability = }}\dfrac{{{\text{Favorable chances}}}}{{{\text{Total chances}}}}$, similar we can find B’s chances of winning a game. Use this concept along with the binomial theorem of probability to reach the answer.

Complete step-by-step answer:

It is given that in a certain game A’s skill is to B’s as 3 to 2……………….. (1)

Using ${\text{Probability = }}\dfrac{{{\text{Favorable chances}}}}{{{\text{Total chances}}}}$……………….. (2)

Let p be the probability of A winning a game.

So, the chance of winning by A (i.e. probability to win by A) using equation (1) and equation (2)

$ \Rightarrow p = \dfrac{3}{{3 + 2}} = \dfrac{3}{5}$.

Let q be the probability of B winning a game. So the probability to win by B using equation (1) and equation (2)

$ \Rightarrow q = \dfrac{2}{{3 + 2}} = \dfrac{2}{5}$

Now we have to find out the chance of A winning 3 games at least out of 5.

Now we use binomial theorem to calculate the required probability.

$ \Rightarrow {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$

Where, n = number of times the game played.

r = number of times A wins.

p = Probability to win by A.

q = Probability to win by B, or losing probability of A.

Now it is given A win at least 3 games (i.e. A win 3 games or 4 games or 5 games)

So, the required probability (${P_A}$) of winning 3 games at least out of 5 is

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

Now as we know ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$

\Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5.4.3!}}{{3!\left( {2 \times 1} \right)}} = 10 \\

\Rightarrow {}^5{C_4} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} = \dfrac{{5.4!}}{{4!\left( 1 \right)}} = 5 \\

\Rightarrow {}^5{C_5} = \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} = \dfrac{{5!}}{{5!\left( {0!} \right)}} = \dfrac{{5!}}{{5!\left( 1 \right)}} = 1 \\

$

So substitute these values in equation (3) we have

${P_A} = 10{\left( {\dfrac{3}{5}} \right)^3}{\left( {\dfrac{2}{5}} \right)^2} + 5{\left( {\dfrac{3}{5}} \right)^4}{\left( {\dfrac{2}{5}} \right)^1} + 1{\left( {\dfrac{3}{5}} \right)^5}{\left( {\dfrac{2}{5}} \right)^0}$

Now simplify the above equation we have,

${P_A} = 10\left( {\dfrac{{27}}{{125}}} \right)\left( {\dfrac{4}{{25}}} \right) + 5\left( {\dfrac{{81}}{{625}}} \right)\left( {\dfrac{2}{5}} \right) + 1\left( {\dfrac{{243}}{{3125}}} \right)\left( 1 \right)$

${P_A} = \dfrac{{1080}}{{3125}} + \dfrac{{810}}{{3125}} + \dfrac{{243}}{{3125}}$

Now add all these terms we have,

${P_A} = \dfrac{{2133}}{{3125}}$

So, this is the required probability of A winning 3 games at least out of 5.

So, this is the required answer.

Note: Whenever we face such types of problems the key point is to take out individual probabilities of winning a specific game then whenever we come across words in questions like at least or at most involving probability then we always have to use the binomial probability theorem. This will help you get on the right track to reach the answer.

Complete step-by-step answer:

It is given that in a certain game A’s skill is to B’s as 3 to 2……………….. (1)

Using ${\text{Probability = }}\dfrac{{{\text{Favorable chances}}}}{{{\text{Total chances}}}}$……………….. (2)

Let p be the probability of A winning a game.

So, the chance of winning by A (i.e. probability to win by A) using equation (1) and equation (2)

$ \Rightarrow p = \dfrac{3}{{3 + 2}} = \dfrac{3}{5}$.

Let q be the probability of B winning a game. So the probability to win by B using equation (1) and equation (2)

$ \Rightarrow q = \dfrac{2}{{3 + 2}} = \dfrac{2}{5}$

Now we have to find out the chance of A winning 3 games at least out of 5.

Now we use binomial theorem to calculate the required probability.

$ \Rightarrow {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$

Where, n = number of times the game played.

r = number of times A wins.

p = Probability to win by A.

q = Probability to win by B, or losing probability of A.

Now it is given A win at least 3 games (i.e. A win 3 games or 4 games or 5 games)

So, the required probability (${P_A}$) of winning 3 games at least out of 5 is

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

Now as we know ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$

\Rightarrow {}^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} = \dfrac{{5.4.3!}}{{3!\left( {2 \times 1} \right)}} = 10 \\

\Rightarrow {}^5{C_4} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} = \dfrac{{5.4!}}{{4!\left( 1 \right)}} = 5 \\

\Rightarrow {}^5{C_5} = \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} = \dfrac{{5!}}{{5!\left( {0!} \right)}} = \dfrac{{5!}}{{5!\left( 1 \right)}} = 1 \\

$

So substitute these values in equation (3) we have

${P_A} = 10{\left( {\dfrac{3}{5}} \right)^3}{\left( {\dfrac{2}{5}} \right)^2} + 5{\left( {\dfrac{3}{5}} \right)^4}{\left( {\dfrac{2}{5}} \right)^1} + 1{\left( {\dfrac{3}{5}} \right)^5}{\left( {\dfrac{2}{5}} \right)^0}$

Now simplify the above equation we have,

${P_A} = 10\left( {\dfrac{{27}}{{125}}} \right)\left( {\dfrac{4}{{25}}} \right) + 5\left( {\dfrac{{81}}{{625}}} \right)\left( {\dfrac{2}{5}} \right) + 1\left( {\dfrac{{243}}{{3125}}} \right)\left( 1 \right)$

${P_A} = \dfrac{{1080}}{{3125}} + \dfrac{{810}}{{3125}} + \dfrac{{243}}{{3125}}$

Now add all these terms we have,

${P_A} = \dfrac{{2133}}{{3125}}$

So, this is the required probability of A winning 3 games at least out of 5.

So, this is the required answer.

Note: Whenever we face such types of problems the key point is to take out individual probabilities of winning a specific game then whenever we come across words in questions like at least or at most involving probability then we always have to use the binomial probability theorem. This will help you get on the right track to reach the answer.

Recently Updated Pages

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE

Let x1x2x3 and x4 be four nonzero real numbers satisfying class 11 maths CBSE

Trending doubts

How many crores make 10 million class 7 maths CBSE

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

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

List out three methods of soil conservation

Write an application to the principal requesting five class 10 english CBSE

What is a collective noun for bees class 10 english CBSE