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Question Number 104708 by byaw last updated on 23/Jul/20

A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn wthout replacement from each bag, find the probability that: i. all the four balls are black ii. exactly one of the four balls is white

$$ \\ $$$$\:\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{12}\:\mathrm{white}\:\mathrm{balls}\: \\ $$$$\mathrm{and}\:\mathrm{8}\:\mathrm{black}\:\mathrm{balls},\:\mathrm{another}\: \\ $$$$\mathrm{contains}\:\mathrm{10}\:\mathrm{white}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{15} \\ $$$$\mathrm{black}\:\mathrm{balls}.\:\mathrm{If}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{drawn} \\ $$$$\mathrm{wthout}\:\mathrm{replacement}\:\mathrm{from}\:\mathrm{each} \\ $$$$\mathrm{bag},\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}:\: \\ $$$$ \\ $$$$\mathrm{i}.\:\mathrm{all}\:\mathrm{the}\:\mathrm{four}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{black}\: \\ $$$$ \\ $$$$\mathrm{ii}.\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four}\:\mathrm{balls}\:\mathrm{is} \\ $$$$\mathrm{white} \\ $$

Answered by som(math1967) last updated on 23/Jul/20

i) All 4 balls are black (c_2 ^8 /c_2 ^(20) )×(c_2 ^(15) /c_2 ^(25) )=((28)/(190))×((105)/(300))=((28)/(950)) ii)(c_2 ^8 /c_2 ^(20) )×((c_1 ^(15) ×c_1 ^(10) )/c_2 ^(25) ) +((c_1 ^8 ×c_1 ^(12) )/c_2 ^(20) )×(c_2 ^(15) /c_2 ^(25) )

$$\left.\mathrm{i}\right)\:\mathrm{All}\:\mathrm{4}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{black} \\ $$$$\frac{\overset{\mathrm{8}} {\mathrm{c}}_{\mathrm{2}} }{\overset{\mathrm{20}} {\mathrm{c}}_{\mathrm{2}} }×\frac{\overset{\mathrm{15}} {\mathrm{c}}_{\mathrm{2}} }{\overset{\mathrm{25}} {\mathrm{c}}_{\mathrm{2}} }=\frac{\mathrm{28}}{\mathrm{190}}×\frac{\mathrm{105}}{\mathrm{300}}=\frac{\mathrm{28}}{\mathrm{950}} \\ $$$$\left.\mathrm{ii}\right)\frac{\overset{\mathrm{8}} {\mathrm{c}}_{\mathrm{2}} }{\overset{\mathrm{20}} {\mathrm{c}}_{\mathrm{2}} }×\frac{\overset{\mathrm{15}} {\mathrm{c}}_{\mathrm{1}} ×\overset{\mathrm{10}} {\mathrm{c}}_{\mathrm{1}} }{\overset{\mathrm{25}} {\mathrm{c}}_{\mathrm{2}} }\:+\frac{\overset{\mathrm{8}} {\mathrm{c}}_{\mathrm{1}} ×\overset{\mathrm{12}} {\mathrm{c}}_{\mathrm{1}} }{\overset{\mathrm{20}} {\mathrm{c}}_{\mathrm{2}} }×\frac{\overset{\mathrm{15}} {\mathrm{c}}_{\mathrm{2}} }{\overset{\mathrm{25}} {\mathrm{c}}_{\mathrm{2}} } \\ $$

Commented by byaw last updated on 23/Jul/20

please help with the question below A box contains 5 white 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining one black ball or red ball

$${please}\:{help}\:{with}\:{the}\:{question} \\ $$$${below} \\ $$$$ \\ $$$$\:\mathrm{A}\:\mathrm{box}\:\mathrm{contains}\:\mathrm{5}\:\mathrm{white}\:\mathrm{3}\:\mathrm{black} \\ $$$$\mathrm{balls}\:\mathrm{and}\:\mathrm{2}\:\mathrm{red}\:\mathrm{balls}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{size}.\:\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{box}\:\mathrm{and}\:\mathrm{then}\:\mathrm{replaced}.\:\mathrm{A} \\ $$$$\mathrm{second}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{then}\:\mathrm{selected}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{obtaining}\:\mathrm{one} \\ $$$$\mathrm{black}\:\mathrm{ball}\:\mathrm{or}\:\mathrm{red}\:\mathrm{ball} \\ $$

Commented by som(math1967) last updated on 24/Jul/20

(WB)or(BW)or(BR)or(RB) (RW)or(WR) (5/(10))×(3/(10)) ×2+(3/(10))×(2/(10))×2+(2/(10))×(5/(10))×2 2×(((31)/(100)))=((31)/(50))

$$\left(\mathrm{WB}\right)\mathrm{or}\left(\mathrm{BW}\right)\mathrm{or}\left(\mathrm{BR}\right)\mathrm{or}\left(\mathrm{RB}\right) \\ $$$$\left(\mathrm{RW}\right)\mathrm{or}\left(\mathrm{WR}\right) \\ $$$$\frac{\mathrm{5}}{\mathrm{10}}×\frac{\mathrm{3}}{\mathrm{10}}\:×\mathrm{2}+\frac{\mathrm{3}}{\mathrm{10}}×\frac{\mathrm{2}}{\mathrm{10}}×\mathrm{2}+\frac{\mathrm{2}}{\mathrm{10}}×\frac{\mathrm{5}}{\mathrm{10}}×\mathrm{2} \\ $$$$\mathrm{2}×\left(\frac{\mathrm{31}}{\mathrm{100}}\right)=\frac{\mathrm{31}}{\mathrm{50}} \\ $$

Commented by 1549442205PVT last updated on 24/Jul/20

this solution is for the question: Find the probality so that two balls drawn different colour sir?Sir looking back once again the question of problem!

$$\mathrm{this}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{for}\:\mathrm{the}\:\mathrm{question}: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{probality}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{two}\:\mathrm{balls}\:\mathrm{drawn}\: \\ $$$$\mathrm{different}\:\mathrm{colour}\:\mathrm{sir}?\mathrm{Sir}\:\mathrm{looking}\:\mathrm{back}\:\mathrm{once}\:\mathrm{again}\: \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{of}\:\mathrm{problem}! \\ $$

Commented by som(math1967) last updated on 24/Jul/20

If one black ball obtain then red ball not obtain then solution is 2×((5/(10))×(3/(10)) +(5/(10))×(2/(10))) 2×(1/4)=(1/2)

$$\mathrm{If}\:\mathrm{one}\:\mathrm{black}\:\mathrm{ball}\:\mathrm{obtain}\:\mathrm{then} \\ $$$$\mathrm{red}\:\mathrm{ball}\:\mathrm{not}\:\mathrm{obtain}\:\mathrm{then} \\ $$$$\mathrm{solution}\:\mathrm{is} \\ $$$$\:\mathrm{2}×\left(\frac{\mathrm{5}}{\mathrm{10}}×\frac{\mathrm{3}}{\mathrm{10}}\:+\frac{\mathrm{5}}{\mathrm{10}}×\frac{\mathrm{2}}{\mathrm{10}}\right) \\ $$$$\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by byaw last updated on 24/Jul/20

Thank you very much

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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