Question Detail

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}

1. 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}

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:

3. A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is

\begin{aligned} \frac{7}{19} \end{aligned}
\begin{aligned} \frac{6}{19} \end{aligned}
\begin{aligned} \frac{5}{19} \end{aligned}
\begin{aligned} \frac{4}{19} \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}

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. 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