Question Detail

From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn.

3/13
1/52
1/4
None of above

Answer: Option A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]

So probability = 12 /52 = 3/13

1. From a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn.

3/13
1/52
1/4
None of above

Answer: Option A

Explanation:

Total number of cases = 52

Total face cards = 12 [favourable cases]

So probability = 12 /52 = 3/13

2. There is a pack of 52 cards and Rohan draws two cards together, what is the probability that one is spade and one is heart ?

11/102
13/102
11/104
11/102

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}

3. Two unbiased coins are tossed. What is probability of getting at most one tail ?

\begin{aligned} \frac{1}{2} \end{aligned}
\begin{aligned} \frac{1}{3} \end{aligned}
\begin{aligned} \frac{3}{2} \end{aligned}
\begin{aligned} \frac{3}{4} \end{aligned}

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

4. Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white

Answer: Option B

Explanation:

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

So probability = 20/30 = 2/3

5. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random. What is the probability that they are not of same colour.

52/55
3/55
41/44
3/44

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}