A bag A contains a white and 3 red balls and
a bag B contains 4 white and 5 red balls. One
ball is drawn at random from

Question

A bag A contains a white and 3 red balls and
a bag B contains 4 white and 5 red balls. One
ball is drawn at random from one of the bags
and
it is found to be red. Find the probability that the
red ball drawn is from bag B.

Josephine · Answer

Given  :     A bag A contains one white and 3 red balls and a bag B contains 4 white and 5 red balls.     Need To Find :   The probability that the red ball drawn is from bag B.    ⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀    ❍ Let's Consider the  A     be the Probability of choosing Bag A ,  B     be the Probability of choosing Bag B   &      R     be the Probability of the ball is red .    Now ,        PROBABILITY       of choosing the Bag  A    or  B     is given by :        qquad  quad  leadsto  sf P(A) =  dfrac{1}{2}          ⠀⠀⠀⠀⠀ AND     ,          qquad  quad  leadsto  sf P(B) =  dfrac{1}{2}          Now ,      PROBABILITY     of choosing a  red ball  from  Bag A  :      ⠀⠀⠀⠀━━ A bag A contains one white and 3 red balls .      qquad  quad  leadsto  sf P  bigg(  dfrac{R}{A}  bigg)  : := : :  dfrac{3}{4}             PROBABILITY    of choosing  red     ball    from  Bag     B     .      ⠀⠀⠀⠀━━ A bag B contains 4 white and 5 red balls.      qquad  quad  leadsto  sf P  bigg(  dfrac{R}{B}  bigg)  : := : :  dfrac{5}{9}           Now ,      By  BAYES'S     THEOREM           :         ⠀⠀⠀⠀⠀⠀The probability that the red ball drawn is from bag B.      qquad  dag  qquad  boxed { pmb{ mathrm{  pink{  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{P (R/ B )  times P(B)}{P(R/B) : : times P(B) : :+ : P(R/A) : times P(A) }}}}}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{P (R/ B )  times P(B)}{P(R/B) : : times P(B) : :+ : P(R/A) : times P(A)}         ⠀⠀⠀⠀⠀⠀  underline { boldsymbol{ star :Now  : By  : Substituting  : the  : known  : Values  ::}}         qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{P (R/ B )  times P(B)}{P(R/B) : : times P(B) : :+ : P(R/A) : times P(A)}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5/ 9 )  times (1 /2 )}{(5/9) : : times (1/2) : :+ : (3/4) : times (1/2)}           qquad  dashrightarrow  sf  : : :  bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5/ 18 )}{(5/9) : : times (1/2) : :+ : (3/4) : times (1/2)}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5/ 18 )}{(5/18) : : : :+ : (3/8) :}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5/ 18 )}{(47/72) : : : : :}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5/  cancel {18} )}{(47/ cancel {72}) : : : : :}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ (5 )}{(47/4) : : : : :}           qquad  dashrightarrow  sf  : : : P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ 20}{47 : : : : :}           qquad  therefore  pmb{ underline{ purple{ :P bigg(  dfrac{B}{R}  bigg)  : := : dfrac{ 20}{47 : : : : :} }} } : bigstar          qquad  therefore  underline{ sf Hence,  : :The  :Probability  :of  :choosing  : red ;ball :from  :Bag  : B  :is  : bf 20/47 }

Kaylee · Answer

Answer:   Probability = 20/47   Step-by-step explanation:     Given:     Bag A contains a white ball and 3 red balls  Bag B contains 4 white balls and 5 red balls       To Find:     Probability that the red ball drawn is from bag B       Solution:    Let E be the event that the ball is red.  Probability of choosing the bag A or B is given by,    sf P(A)=P(B)= dfrac{1}{2}   Given that bag A contains 3 red balls and 1 white ball.  Hence the probability of drawing a red ball from bag A is given by,    sf P(E/A)= dfrac{3}{4}   Also by given bag B contains 5 red balls and 4 white balls.  Therefore probability of drawing a red ball from bag B is given by,    sf P(E/B)= dfrac{5}{9}     Now we have to find the probability that the red ball was drawn from the bag B, that is P(B/E)  By Bayes theorem,    sf P(B/E)= dfrac{P(E/B) times P(B)}{P(E/B) times P(B)+P(E/A) times P(A)}   Substituting the values we get,    sf P(B/E)= dfrac{5/9  times 1/2}{5/9 times 1/2+3/4 times 1/2}     sf  implies  dfrac{5/18}{5/18+3/8}     sf  implies  dfrac{5/18}{47/72}     sf  implies  dfrac{20}{47}   Hence   the probability that the ball drawn is from bag B is 20/47.

Everleigh

Q.3 Draw the following diagrams –
a . any five fat giving food -

A4.He was sure she would not give him away. (Use modal auxilliary showing ability)

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A bag A contains a white and 3 red balls anda bag B contains 4 white and 5 red balls. Oneball is drawn at random from