Question Detail
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
1. 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
2. Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?
- 1/3
- 1/6
- 1/2
- 1/8
Answer: Option C
Explanation:
Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]
Favoured cases are = [TTH, THT, HTT, TTT] = 4
So required probability = 4/8 = 1/2
3. 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
4. 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}
5. There is a pack of 52 cards and Rohan draws two cards together, what is the probability that one is spade and one is heart ?
- 11/102
- 13/102
- 11/104
- 11/102
Answer: Option B
Explanation:
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
Let sample space be S
\begin{aligned}
\text{then, n(S) = }^{52} C _2 \\
=> \frac{52 \times 51}{ 2 \times 1} = 1326 \\
\text {let E be event of getting 1 spade and 1 heart} \\
\text{So, n(E) = ways of getting 1 spade or 1 heart out of 13} \\
= ^{13}C_1 \times ^{13}C_1 \\
= 13 \times 13 \\
= 169 \\
\text{So, p(E) = }\frac{n(E)}{n(S)} \\
= \frac{169}{1326} = \frac{13}{102}
\end{aligned}