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

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 A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]



So probability = 12 /52 = 3/13

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
$$\begin{aligned} ^{16}C_2 = \frac{16*15}{2*1} = 120 \\ \text{total cases = } \\ ^{20}C_2 = \frac{20*19}{2*1} = 190 \\ \text{probability = } \frac{120}{190} = \frac{12}{19} \\ \text{P of at least one defective = } 1- \frac{12}{19} \\ =\frac{7}{19} \end{aligned}$$

Answer: Option B

Explanation:

Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

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