Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?

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%

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

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

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

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}