Question Detail

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

1. 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

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

1/13
2/13
1/26
1/52

Answer: Option C

Explanation:

Total number of cases = 52
Favourable cases = 2

Probability = 2/56 = 1/26

3. 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

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 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}