13Find the probability that a number selected at random from 1 to 40 is divisible by either 4 or 6. A1/10 B5/10 C2/10 D3/10

Answer: D

Explanation:
We have to use addition formula here.
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
From 1 to 40, Numbers divisible by 4 $= 9$ ($\because$ 4, 8, 12, 16, 20, 24, 28, 32, 36, 40)
Numbers divisible by 6 $= 6$ ($\because$ 6, 12, 18, 24, 30, 36)
Numbers divisible by both 4 and 6 are 3 ($\because$ $\text{LCM}(4, 6) = 12 \;\text{and 12 multiples} = 12, 24, 36$)
A number selected at random from 1 to 40 is divisible by either 4 or 6 $= 9+6-3 = 12$
Total outcomes $= 40$
$\therefore$ Required probability $ = \dfrac{{12}}{{40}} = \dfrac{3}{{10}}$

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14A box contain 14 apples out of which 5 are spoilt. If 5 apples are chosen at random, find the probability that all the 5 apples are spoilt. A1/12225 B1/12852 C1/18018 D1/12552

Answer: C

Explanation:
Out of 14, 5 are spoilt and 9 are good
$\therefore$ Required probability $ = \dfrac{{{}^5{C_5}}}{{{}^{14}{C_5}}}$ $ = \dfrac{1}{{\left( {\dfrac{{14 \times 13 \times 12 \times 11 \times 10}}{{5 \times 4 \times 3 \times 2 \times 1}}} \right)}}$ $ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{14 \times 13 \times 12 \times 11 \times 10}}$$ = \dfrac{1}{{18018}}$

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15A box contain 12 batteries out of which 7 are defective. 3 batteries are drawn from box at random. What is the probability that there is at least one defective battery? A24/44 B22/44 C21/22 D25/44

Answer: C

Explanation:
Number of good batteries $= 12-7 = 5$
Total number of outcomes $={{}^{12}{C_3}} = 220$
Number of outcomes that at least one battery drawn out of three batteries is defective = 1 - (all are good)
$\therefore$ $1 - \dfrac{{{}^5{C_3}}}{{{}^{12}{C_3}}}$ $ = 1 - \dfrac{{10}}{{220}}$ $=\dfrac{{21}}{{22}}$

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168 girls sit in a row randomly. What is the probability that 4 particular girls always sit together? A1/14 B5/14 C7/14 D4/14

Answer: A

Explanation:
Let us group those four girls into $X$.
Now we have $G,G,G,G,\boxed{G,G,G,G}_X$
Number of ways of arranging 5 above objects $=5!$
Now, the four girls may arrange themselves in $4!$ ways.
$\therefore$ In a row of 8 girls, 4 particular girls always sitting together can be arranged in $5!\times 4!$ ways.
Eight persons can be seated in a row in $8!$ ways.
$\therefore$ Required probability $= \dfrac{{5! \times 4!}}{{8!}} = \dfrac{1}{{14}}$

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17The probability that A speaks truth is 5/12 and the probability that B speak truth is 5/14. What is the probability that both speak falsehood while making a statement? A2/8 B1/8 C3/8 D5/8

Answer: C

Explanation:
The probability that A speaks falsehood $ = 1 - \dfrac{5}{{12}} = \dfrac{7}{{12}}$
The probability that B speaks falsehood $ = 1 - \dfrac{5}{{14}} = \dfrac{9}{{14}}$
$\therefore$ Required probability $= \dfrac{7}{{12}} \times \dfrac{9}{{14}} = \dfrac{3}{8}$

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18Bag $P$ contain $6$ Red marbles and $5$ blue marbles while Bag $Q$ contain $5$ Red marbles and $6$ blue marbles. Two marbles, one each from the two bags are drawn at random. What is the probability that both the marbles are of same colour? A9/121 B95/121 C60/121 D98/121

Answer: C

Explanation:
Required probability = Probability of drawing red from both bags + Probability of drawing blue from both bags
$\therefore$ $\dfrac{{{}^6{C_1}}}{{{}^{11}{C_1}}} \times \dfrac{{{}^5{C_1}}}{{{}^{11}{C_1}}} + \dfrac{{{}^5{C_1}}}{{{}^{11}{C_1}}} \times \dfrac{{{}^6{C_1}}}{{{}^{11}{C_1}}}$ $ = \dfrac{{60}}{{121}}$