Math, asked by shilpidhali, 11 months ago

A and B played a game in which their chances of winning are in the ratio 3:2. A's chances of winning 3 games out of 5 game

Answers

Answered by pulakmath007
8

SOLUTION

TO DETERMINE

A and B play a game in which their chances of winning are in the ratio 3:2. Find A's chance of winning of at least three games out of the five games played

CONCEPT TO BE IMPLEMENTED

If a trial is repeated n times and p is the probability of a success and q that of failure then the probability of r successes is

 \displaystyle \sf{  \sf{P(X=r) =  \:  \: }\large{ {}^{n} C_r}\:  {p}^{r}  \:  \:  {q}^{n - r} } \:  \:  \:  \:  \: where \: q \:  = 1 - p

EVALUATION

Here it is given that A and B played a game in which their chances of winning are in the ratio 3 : 2

So p = Probability of A's win

 \displaystyle \sf{ =  \frac{3}{3 + 2} }

 \displaystyle \sf{ =  \frac{3}{5} }

 \displaystyle \sf{ =  0.6}

So q = 1 - p = 1 - 0.6 = 0.4

Now we have to find the A's chances of winning at least 3 games out of 5 game

Thus n = 5

Hence the required probability

\displaystyle \sf{  = P(X \geqslant 3)}

\displaystyle \sf{  = P(X  =  3) +P(X  =  4) +P(X  =  5)  }

\displaystyle \sf{ =   {}^{5} C_3 {(0.6)}^{ 3}    {(0.4)}^{2}  + {}^{5} C_4 {(0.6)}^{ 4}    {(0.4)}^{1}  +{}^{5} C_5 {(0.6)}^{ 5}    {(0.4)}^{0}  }

= 0.3456 + 0.2592 + 0.0777

= 0.6825

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Answered by varshinipriya42309
1

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