A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident

Answer: Option C

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26

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 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:

Total cases = 10 + 20 = 30
Favourable cases = 20

So probability = 20/30 = 2/3

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