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. Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white

Answer: Option B

Explanation:

Total cases = 10 + 20 = 30
Favourable cases = 20

So probability = 20/30 = 2/3

2. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

Answer: Option A

Explanation:

Total number of cases = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So Probability = 27/36 = 3/4

3. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is

1/13
2/13
1/26
1/52

Answer: Option C

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26

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. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random. What is the probability that they are not of same colour.

52/55
3/55
41/44
3/44

Answer: Option C

Explanation:

\begin{aligned}
\text{Total cases =} ^{12}C_3 \\
= \frac{12*11*10}{3*2*1} = 220 \\
\text{Total cases of drawing same colour =} \\
^{5}C_3 + ^{4}C_3 + ^{3}C_3 \\
\frac{5*4}{2*1} + 4 + 1 = 15 \\

\text{Probability of same colur =} = \frac{15}{220}\\
= \frac{3}{44} \\

\text{Probability of not same colur =} \\
1-\frac{3}{44}\\ = \frac{41}{44}

\end{aligned}