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Q:

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: D) 1/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is $7 C_{5}$ .

So,required probability = $7 C_{5}$ / $10 C_{5}$ = 1/12.

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Q:

A box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains exactly one defective bulb.

A) 5/12	B) 7/12
C) 3/14	D) 1/12

Answer & Explanation Answer: A) 5/12

Explanation:

Total number of elementary events = $10 C_{5}$

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is $3 C_{1} * 7 C_{4}$

So,required probability = $3 C_{1} * 7 C_{4}$ / $10 C_{5}$ = 5/12.

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Q:

A bag contains 50 tickets numbered 1,2,3,4......50 of which five are drawn at random and arranged in ascending order of magnitude.Find the probability that third drawn ticket is equal to 30.

A) 551/15134	B) 1/2
C) 552/15379	D) 1/9

Answer & Explanation Answer: A) 551/15134

Explanation:

Total number of elementary events = $50 C_{5}$
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = $29 C_{2}$ ways and remaining two in $20 C_{2}$ ways.

Therfore,favourable number of events = $29 C_{2} * 20 C_{2}$

Hence,required probability = $29 C_{2} * 20 C_{2}$ $/ 50 C_{5}$ =551 / 15134

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Q:

A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

A) 49	B) 50
C) 53.5	D) 55

Answer & Explanation Answer: A) 49

Explanation:

Area of the wet surface = [2(lb + bh + lh) - lb]

= 2(bh + lh) + lb

= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] $m^{2}$

= 49 $m^{2}$

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Filed Under: Volume and Surface Area

Q:

What is the probability that a leap year selected at random, will contain 53 sundays?

A) 1/7	B) 1/3
C) 2/7	D) 4/7

Answer & Explanation Answer: C) 2/7

Explanation:

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

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Q:

Find the probability that it is either white or red ball when,one ball is drawn at random in a bag that contains 3 black,4 white and 5 red balls

A) 0.5	B) 0.75
C) 0.25	D) 0.8

Answer & Explanation Answer: B) 0.75

Explanation:

P(red ball) = 5/12

P(white ball) = 4/12

P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75

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Q:

A bag contains 3 black, 4 white and 5 red balls. One ball is drawn at random. Find the probability that it is either black or red ball:

A) 2/3	B) 1/4
C) 5/12	D) 1/2

Answer & Explanation Answer: A) 2/3

Explanation:

P(black ball)=3/12

P(red ball)=5/12

P(black or red)=3/12+5/12=2/3

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Q:

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A) 3/7	B) 4/7
C) 1/8	D) 3/4

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