A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is

Answer: Option A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]



So probability = 12 /52 = 3/13

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 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 C

Explanation:

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

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
$$\begin{aligned} ^{16}C_2 = \frac{16*15}{2*1} = 120 \\ \text{total cases = } \\ ^{20}C_2 = \frac{20*19}{2*1} = 190 \\ \text{probability = } \frac{120}{190} = \frac{12}{19} \\ \text{P of at least one defective = } 1- \frac{12}{19} \\ =\frac{7}{19} \end{aligned}$$