Question Detail

A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

$\begin{aligned} \frac{7}{19} \end{aligned}$
$\begin{aligned} \frac{6}{19} \end{aligned}$
$\begin{aligned} \frac{5}{19} \end{aligned}$
$\begin{aligned} \frac{4}{19} \end{aligned}$

Answer: Option A

Explanation:

Please remember that Maximum portability is 1.

So we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs.

So here we go,
Total cases of non defective bulbs

$\begin{aligned} ^{16}C_2 = \frac{16*15}{2*1} = 120 \\ \text{total cases = } \\ ^{20}C_2 = \frac{20*19}{2*1} = 190 \\ \text{probability = } \frac{120}{190} = \frac{12}{19} \\ \text{P of at least one defective = } 1- \frac{12}{19} \\ =\frac{7}{19} \end{aligned}$

1. From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn.

3/13
1/52
1/4
None of above

Answer: Option A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]

So probability = 12 /52 = 3/13

2. In a throw of coin what is the probability of getting head.

Answer: Option C

Explanation:

Total cases = [H,T] - 2
Favourable cases = [H] -1
So probability of getting head = 1/2

3. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

2/121
2/221
1/221
1/13

Answer: Option C

Explanation:

$\begin{aligned} \text{Total cases =} ^{52}C_2 \\ = \frac{52*51}{2*1} = 1326 \\ \text{Total King cases =} ^{4}C_2 \\ = \frac{4*3}{2*1} = 6 \\ \text{Probability =} = \frac{6}{1326}\\ = \frac{1}{221} \\ \end{aligned}$

4. Two unbiased coins are tossed. What is probability of getting at most one tail ?

$\begin{aligned} \frac{1}{2} \end{aligned}$
$\begin{aligned} \frac{1}{3} \end{aligned}$
$\begin{aligned} \frac{3}{2} \end{aligned}$
$\begin{aligned} \frac{3}{4} \end{aligned}$

Answer: Option D

Explanation:

Total 4 cases = [HH, TT, TH, HT]
Favourable cases = [HH, TH, HT]
Please note we need atmost one tail, not atleast one tail.

So probability = 3/4

5. What is the probability of getting a sum 9 from two throws of dice.

1/3
1/9
1/12
2/9

Answer: Option B

Explanation:

Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

anu 11 years ago

why we hve taken combination ist ie 16c2

Question Detail

A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

1. From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn.

2. In a throw of coin what is the probability of getting head.

3. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

4. Two unbiased coins are tossed. What is probability of getting at most one tail ?

5. What is the probability of getting a sum 9 from two throws of dice.

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anu 11 years ago

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Question Detail

A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

1. From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn.

2. In a throw of coin what is the probability of getting head.

3. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

4. Two unbiased coins are tossed. What is probability of getting at most one tail ?

5. What is the probability of getting a sum 9 from two throws of dice.

Thanks ! Your comment will be approved shortly !

anu 11 years ago

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