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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. Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white
- 1
- 2/3
- 1/3
- 4/3
Answer: Option B
Explanation:
Total cases = 10 + 20 = 30
Favourable cases = 20
So probability = 20/30 = 2/3
2. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?
- 2/3
- 8/21
- 3/7
- 9/22
Answer: Option B
Explanation:
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
3. In a throw of coin what is the probability of getting tails.
- 1
- 2
- 1/2
- 0
Answer: Option C
Explanation:
Total cases = [H,T] - 2
Favourable cases = [T] -1
So probability of getting tails = 1/2
4. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?
- 3/4
- 1/4
- 7/4
- 1/2
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
5. A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident
- 30%
- 35%
- 40%
- 45%
Answer: Option B
Explanation:
Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = 1-3/4 = 1/4
P(B-lie) = 1-4/5 = 1/5
Now
A and B contradict each other =
[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
[Please note that we are adding at the place of OR]
= (3/5*1/5) + (1/4*4/5) = 7/20
= (7/20 * 100) % = 35%
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