In a throw of coin what is the probability of getting 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 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

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

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.

Answer: Option C

Explanation:

$$\begin{aligned} \text{Total cases =} ^{12}C_3 \\ = \frac{12*11*10}{3*2*1} = 220 \\ \text{Total cases of drawing same colour =} \\ ^{5}C_3 + ^{4}C_3 + ^{3}C_3 \\ \frac{5*4}{2*1} + 4 + 1 = 15 \\ \text{Probability of same colur =} = \frac{15}{220}\\ = \frac{3}{44} \\ \text{Probability of not same colur =} \\ 1-\frac{3}{44}\\ = \frac{41}{44} \end{aligned}$$

Answer: Option C

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26