In a throw of coin what is the probability of getting head.

Answer: Option C

Explanation:

Total cases = [H,T] - 2
Favourable cases = [H] -1
So probability of getting head = 1/2

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 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:

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 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