Match the items of List I with items of List II and choose the correct option. COLUMN – I COLUMN – IIi)7 identical white balls and 3 identical black balls are placed in a row at random.The probability that no two black balls are adjacent isP)511ii)4 gentlemen and 4 ladies take seats at random round a table. The probability that they are sitting alternately isQ)116iii)10 different books and 2 different pens are given to 3 boys, so that each gets equal number of things. The probability that the same boy does not receive both the pens, isR)715iv)A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses isS)135 T)316

(a) n (S)=10 !(7 !) (3 !)n(E)=8C3=8 !(3 !) (5 !), because there are 8 places for 3 black balls. ∴ P(E)=8!(3!) (5!)10!(7!) (3!)=(8!) (7!)(10!) (5!)=7.610.9=715b) n(S)=7 !, n(E) = (3!) × (4!) ( ∵ after making 4 gentlemen sit in 3! ways, 4 ladies can sit in 4! ways in between the gentlemen) P(E) = (3!) × (4!)7!=67×6×5=135c) n(S)=12C4×8C4×4C4n(E)= n(S) − the number of ways in which one boy gets both the pens. =n(S)−10C2×8C4×4C4×(3 !)∴ P(E)=1− 10C2×8C4×4C4×(3!) 12C4×8C4×4C4=511d) Required probability= P (E E E E E¯)+P(E¯ E¯ E¯ E¯ E)={P(E)}4 . P(E¯)+{P(E¯)}4 . P(E)=116

Match the items of List I with items of List II and choose the correct option. COLUMN – I COLUMN – IIi)7 identical white balls and 3 identical black balls are placed in a row at random.The probability that no two black balls are adjacent isP)511ii)4 gentlemen and 4 ladies take seats at random round a table. The probability that they are sitting alternately isQ)116iii)10 different books and 2 different pens are given to 3 boys, so that each gets equal number of things. The probability that the same boy does not receive both the pens, isR)715iv)A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses isS)135 T)316

(a) n (S)=10 !(7 !) (3 !)n(E)=8C3=8 !(3 !) (5 !), because there are 8 places for 3 black balls. ∴ P(E)=8!(3!) (5!)10!(7!) (3!)=(8!) (7!)(10!) (5!)=7.610.9=715b) n(S)=7 !, n(E) = (3!) × (4!) ( ∵ after making 4 gentlemen sit in 3! ways, 4 ladies can sit in 4! ways in between the gentlemen) P(E) = (3!) × (4!)7!=67×6×5=135c) n(S)=12C4×8C4×4C4n(E)= n(S) − the number of ways in which one boy gets both the pens. =n(S)−10C2×8C4×4C4×(3 !)∴ P(E)=1− 10C2×8C4×4C4×(3!) 12C4×8C4×4C4=511d) Required probability= P (E E E E E¯)+P(E¯ E¯ E¯ E¯ E)={P(E)}4 . P(E¯)+{P(E¯)}4 . P(E)=116

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Match the items of List I with items of List II and choose the correct option.

COLUMN – I COLUMN – II
i) 7 identical white balls and 3 identical black balls are placed in a row at random.The probability that no two black balls are adjacent is P) $\frac{5}{11}$
ii) 4 gentlemen and 4 ladies take seats at random round a table. The probability that they are sitting alternately is Q) $\frac{1}{16}$
iii) 10 different books and 2 different pens are given to 3 boys, so that each gets equal number of things. The probability that the same boy does not receive both the pens, is R) $\frac{7}{15}$
iv) A fair coin is tossed repeatedly. The probability of getting a result in the fifth toss different from those obtained in the first four tosses is S) $\frac{1}{35}$
T) $\frac{3}{16}$

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i-R, ii-S, iii-P, iv-Q

i-S, ii-R, iii-Q, iv-P

i-P, ii-Q, iii-R, iv-S

i-P, ii-S, iii-R, iv-Q

answer is B.

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Detailed Solution

(a) $n (S) = \frac{10!}{(7!) (3!)}$
$n (E) =^{8} C_{3} = \frac{8!}{(3!) (5!)}$ , because there are 8 places for 3 black balls.
$∴ P (E) = \frac{\frac{8!}{(3!) (5!)}}{\frac{10!}{(7!) (3!)}} = \frac{(8!) (7!)}{(10!) (5!)} = \frac{7.6}{10.9} = \frac{7}{15}$

b) $n (S) = 7!, n (E) = (3!) \times (4!)$
( $∵$ after making 4 gentlemen sit in 3! ways, 4 ladies can sit in 4! ways in between the gentlemen)
$P (E) = \frac{(3!) \times (4!)}{7!} = \frac{6}{7 \times 6 \times 5} = \frac{1}{35}$

c) $n (S) =^{12} C_{4} \times^{8} C_{4} \times^{4} C_{4}$
$n (E) = n (S) -$ the number of ways in which one boy gets both the pens.
$= n (S) -^{10} C_{2} \times^{8} C_{4} \times^{4} C_{4} \times (3!)$

$∴ P (E) = 1 - \frac{^{10} C_{2} \times^{8} C_{4} \times^{4} C_{4} \times (3!)}{^{12} C_{4} \times^{8} C_{4} \times^{4} C_{4}} = \frac{5}{11}$

d) Required probability= $P (E E E E \bar{E}) + P (\bar{E} \bar{E} \bar{E} \bar{E} E)$

$= {P (E)}^{4} . P (\bar{E}) + {P (\bar{E})}^{4} . P (E) = \frac{1}{16}$

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