Question Detail

In a throw of dice what is the probability of getting number greater than 5

Answer: Option D

Explanation:

Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]

So probability = 1/6

1. 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

2. 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:

$\begin{aligned} \text{Total cases =} ^{12}C_3 \\ = \frac{12*11*10}{3*2*1} = 220 \\ \text{Total cases of drawing same colour =} \\ ^{5}C_3 + ^{4}C_3 + ^{3}C_3 \\ \frac{5*4}{2*1} + 4 + 1 = 15 \\ \text{Probability of same colur =} = \frac{15}{220}\\ = \frac{3}{44} \\ \text{Probability of not same colur =} \\ 1-\frac{3}{44}\\ = \frac{41}{44} \end{aligned}$

3. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?

2/3
8/21
3/7
9/22

Answer: Option B

Explanation:

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.

4. 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

5. 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
$\begin{aligned} \text{then, n(S) = }^{52} C _2 \\ => \frac{52 \times 51}{ 2 \times 1} = 1326 \\ \text {let E be event of getting 1 spade and 1 heart} \\ \text{So, n(E) = ways of getting 1 spade or 1 heart out of 13} \\ = ^{13}C_1 \times ^{13}C_1 \\ = 13 \times 13 \\ = 169 \\ \text{So, p(E) = }\frac{n(E)}{n(S)} \\ = \frac{169}{1326} = \frac{13}{102} \end{aligned}$