Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white

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 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

Answer: Option A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]



So probability = 12 /52 = 3/13

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
$$\begin{aligned} \text{then, n(S) = }^{52} C _2 \\ => \frac{52 \times 51}{ 2 \times 1} = 1326 \\ \text {let E be event of getting 1 spade and 1 heart} \\ \text{So, n(E) = ways of getting 1 spade or 1 heart out of 13} \\ = ^{13}C_1 \times ^{13}C_1 \\ = 13 \times 13 \\ = 169 \\ \text{So, p(E) = }\frac{n(E)}{n(S)} \\ = \frac{169}{1326} = \frac{13}{102} \end{aligned}$$