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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
- \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. In a throw of dice what is the probability of getting number greater than 5
- 1/2
- 1/3
- 1/5
- 1/6
Answer: Option D
Explanation:
Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]
So probability = 1/6
2. In a throw of coin what is the probability of getting head.
- 1
- 2
- 1/2
- 0
Answer: Option C
Explanation:
Total cases = [H,T] - 2
Favourable cases = [H] -1
So probability of getting head = 1/2
3. 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}
4. 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
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}
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