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Match the following Column I data with corresponding probabilities in Column II
Column –I Column –II
i) A box contain 6 balls of unknown colours.3 balls are drawn and they are found to the white. Then the probability that all the 6 balls are white is p) 31/42
ii) If a 4 digit number is formed at random using 0,1,2,3,4,5 Then the probability that it is divisible by 6 is q) 4/7
iii) A purse contains 4 one rupee coins and 6 ten paise coins. If 5 coins are selected at random, then the Probability that the sum on the exceed Rs. 2.25 is r) 1/3
iv) Four digit numbers are formed with 1,2,3,4,5,6 with repetition Then the probability that they are divisible by 3 is s) 13/75

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i) p; ii) q; iii) s; iv) r

i) p; ii) s; iii) q; iv) r

i) q; ii) s; iii) p; iv) r

i) p; ii) q; iii) r; iv) s

answer is D.

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Detailed Solution

(a) Required Probability $= \frac{^{6} C_{3}}{^{3} C_{3} +^{4} C_{3} +^{5} C_{3} +^{6} C_{3}} = \frac{4}{7}$
(b) Using $1, 2, 4, 5 = 3! \times 2$
Using $0, 3, 4, 5 = 3! + (3! - 2!) = 10$
Using $0, 2, 3, 4 = 3! + 2 (3! - 2!) = 14$
Using $0, 1, 3, 5 = 3! = 6$
Using $0, 1, 2, 3 = 3! + (3! - 2!) = 10$

----------------
$n (E) = 52$
----------------
$P (E) = \frac{52}{^{6} P_{4} -^{5} P_{3}} = \frac{13}{75}$
(c)
Rupee (4) Paise (6)
2 3
3 2
4 1
Required Probability $= \frac{^{4} C_{3} .^{6} C_{2} +^{4} C_{2} .^{6} C_{3} +^{4} C_{4} .^{6} C_{1}}{^{10} C_{5}}$

(d) $n (E) = 6 \times 6 \times 6 \times 2$

$n (s) = 6 \times 6 \times 6 \times 6$

$P (E) = \frac{1}{3}$

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