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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}

Similar Questions :

1. Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?

  • 1/3
  • 1/6
  • 1/2
  • 1/8

Answer: Option C

Explanation:

Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]

Favoured cases are = [TTH, THT, HTT, TTT] = 4

So required probability = 4/8 = 1/2

2. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?

  • 3/4
  • 1/4
  • 7/4
  • 1/2

Answer: Option A

Explanation:

Total number of cases = 6*6 = 36

Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27

So Probability = 27/36 = 3/4

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. What is the probability of getting a sum 9 from two throws of dice.

  • 1/3
  • 1/9
  • 1/12
  • 2/9

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

5. 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.

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