In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
Answer: C
There are \( 10 + 25 = 35 \) tickets in the lottery.
\[ P = \frac{10}{35} = \frac{2}{7} \]
The denominator is the total number of tickets, not the number of blanks - using 25 would give \( \tfrac{2}{5} \).
Please log in or register to report this question.
In a set of 30 game cards, 17 are white and rest are green. 4 white and 5 green are marked IMPORTANT. If a card is chosen randomly from this set, what is the possibility of choosing a green card or an ‘IMPORTANT’ card?
Answer: C
Of the 30 cards, 17 are white and 13 green. Nine are marked IMPORTANT: 4 white and 5 green.
Adding the two counts directly would double-count the 5 green IMPORTANT cards, so use the addition rule:
\[ P(G \cup I) = P(G) + P(I) - P(G \cap I) \]
\[ = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30} \]
Equivalently, count the 13 green cards plus the 4 white IMPORTANT ones: \( \tfrac{17}{30} \).
Please log in or register to report this question.
The probability of getting exactly one head in tossing a pair of coins is
Answer: C
Two coins give four outcomes: HH, HT, TH, TT. Exactly one head occurs in HT and TH.
\[ P = \frac{2}{4} = \frac{1}{2} \]
HT and TH are distinct outcomes because the coins are distinguishable; treating them as one would give \( \tfrac{1}{3} \).
Please log in or register to report this question.
Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is -.
Answer: D
Among the first 20 natural numbers:
\[ \text{even} = 10, \qquad \text{prime} = 8 \ (2, 3, 5, 7, 11, 13, 17, 19) \]
Only 2 is both, so
\[ 10 + 8 - 1 = 17 \quad\Longrightarrow\quad P = \frac{17}{20} \]
2 is the only even prime, which is exactly why the overlap here is a single number.
Please log in or register to report this question.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
Answer: D
There are four kings in the pack.
\[ P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \]
Equivalently \( \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{1}{221} \), the second fraction reflecting that the first card is not replaced.
Please log in or register to report this question.
A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Answer: C
There is exactly one queen of clubs and one king of hearts, and no card is both.
\[ P = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26} \]
The question names specific cards, not whole ranks, so the count is 2 rather than 8.
Please log in or register to report this question.
If three coins are tossed simultaneously, than the probability of getting at least two heads, is
Answer: C
Three coins give \( 2^{3} = 8 \) equally likely outcomes.
Those with at least two heads are
\[ \text{HHT},\ \text{HTH},\ \text{THH},\ \text{HHH} \]
\[ P = \frac{4}{8} = \frac{1}{2} \]
"At least two" includes all three, so the HHH case must not be left out; counting only exactly two would give \( \tfrac{3}{8} \).
Please log in or register to report this question.
The collection of one or more outcomes from an experiment is called
Answer: B
An event is any subset of the sample space - one outcome or a collection of them.
Rolling an even number, for example, is the event \( \{2, 4, 6\} \).
Probability is the number assigned to an event, and a random variable is a function from outcomes to numbers - neither is the collection itself.
Please log in or register to report this question.
A box has 5 black and 3 green shirts. One shirt is picked randomly and put in another box. The second box has 3 black and 5 green shirts. Now a shirt is picked from second box. What is the probability of it being a black shirt?
Answer: B
The colour of the transferred shirt is unknown, so split into the two cases and add them.
A black shirt is moved (probability \( \tfrac{5}{8} \)). The second box then has 4 black out of 9:
\[ \frac{5}{8} \times \frac{4}{9} = \frac{20}{72} \]
A green shirt is moved (probability \( \tfrac{3}{8} \)). The second box still has 3 black out of 9:
\[ \frac{3}{8} \times \frac{3}{9} = \frac{9}{72} \]
\[ P = \frac{20}{72} + \frac{9}{72} = \frac{29}{72} \]
The second box holds 9 shirts after the transfer, not 8 - the moved shirt is added to it.
Please log in or register to report this question.
A bag contains 6 black, 7 red and 2 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is black or white.
Answer: A
Total number of balls = 15
Number of balls that are either black or white = 8
Hence the number of favourable outcomes of ball drawn being black or white is 8.
We know that, Probability of an event E, P (E) =number of favourable outcomes/total number of outcomes
So, the required probability is 8/15.
Please log in or register to report this question.
