Ravi and Rashmi each hold 2 red cards and 2 b | vedclass

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Question

(A) Initially,Ravi has ${2R, 2B}$ and Rashmi has ${2R, 2B}$. Total cards with each is $4$.Step $1$: Ravi picks a card from Rashmi. Rashmi now has $3$

Vedclass · Accepted Answer

$p \leq 5 \%$

Vedclass · Answer

(A) Initially,Ravi has ${2R, 2B}$ and Rashmi has ${2R, 2B}$. Total cards with each is $4$.Step $1$: Ravi picks a card from Rashmi. Rashmi now has $3$ cards. Then Rashmi picks a card from Ravi. Ravi now has $4$ cards again.To have all $4$ cards of the same colour,Ravi must end up with $4$ Red cards and Rashmi with $4$ Black cards,or vice versa.Case $1$: Ravi ends with $4$ Red cards and Rashmi with $4$ Black cards.In the first exchange,Ravi must pick a Red card from Rashmi (prob $\frac{2}{4}$) and Rashmi must pick a Black card from Ravi (prob $\frac{2}{4}$). After this,Ravi has ${2R, 1B, 1R} = {3R, 1B}$ and Rashmi has ${1R, 2B, 1B} = {1R, 3B}$.In the second exchange,Ravi must pick the remaining Red card from Rashmi (prob $\frac{1}{4}$) and Rashmi must pick the remaining Black card from Ravi (prob $\frac{1}{4}$). Probability of Case $1$ = $(\frac{2}{4} 	imes \frac{2}{4}) 	imes (\frac{1}{4} 	imes \frac{1}{4}) = \frac{4}{16} 	imes \frac{1}{16} = \frac{4}{256} = \frac{1}{64}$.Case $2$: Ravi ends with $4$ Black cards and Rashmi with $4$ Red cards.By symmetry,the probability is also $\frac{1}{64}$.Total probability $p = \frac{1}{64} + \frac{1}{64} = \frac{2}{64} = \frac{1}{32} = 0.03125 = 3.125 \%$.Since $3.125 \% \leq 5 \%$,the correct option is $A$.

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