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:

\begin{aligned}
\text{Total cases =} ^{52}C_2 \\
= \frac{52*51}{2*1} = 1326 \\
\text{Total King cases =} ^{4}C_2 \\
= \frac{4*3}{2*1} = 6 \\

\text{Probability =} = \frac{6}{1326}\\
= \frac{1}{221} \\
\end{aligned}

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:

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 cases = 10 + 20 = 30
Favourable cases = 20

So probability = 20/30 = 2/3

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