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. From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

2/121
2/221
1/221
1/13

Answer: Option C

Explanation:

\begin{aligned}
\text{Total cases =} ^{52}C_2 \\
= \frac{52*51}{2*1} = 1326 \\
\text{Total King cases =} ^{4}C_2 \\
= \frac{4*3}{2*1} = 6 \\

\text{Probability =} = \frac{6}{1326}\\
= \frac{1}{221} \\
\end{aligned}

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

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. 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

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}