Probability (NCO)

There are 16 elements in the sample space. Then, the event of at least one head is the set of all elements but TTTT is not included.  Hence, there are 15 elements left in this set. Hence, the probability will be 15/16. So, it is the correct answer.

Here, the sample space is 60. A player hits sixes 12 times out of 60 balls. This means that he/she does not hit any six in 60 – 12 = 48 balls. Hence, the probability that a player does not hit any six will be 48/60.

In a die, there are only six numbers, i.e. 1, 2, 3, 4, 5, 6. A prime number is a number which can be divided by itself and 1. Now, find a prime number that is greater than or equal to 5, which is only number five. Hence, the probability will be 1/6. Hence, it is the correct answer.

The sample space for this event is 16 (2 * 2 * 2 * 2). No tails will occur only when an event (HHHH) will occur. Hence, the probability of getting no tails is 1/16.

It is the correct answer. In a die, there are only six numbers – 1, 2, 3, 4, 5, and 6. Various events of getting sum of dice as seven when rolling a pair of dice will be as follows:
{(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}

In a bag, there are total 12 (5 + 4 + 3) balls. For getting 3 red balls, the probability is 3/12 and for getting 4 yellow balls, the probability is 4/12. Hence, the probability of getting red and green balls by using the product principle, is (3/12) * (4/12) = 1/12.

In a bag, there are total 12 (5 + 4 + 3) balls. For getting 3 red balls, the probability is 3/12 and for getting 4 green balls, the probability is 4/12. Hence, getting either a red ball or a green ball by addition principle, the probability will be (3/12) + (4/12) = 7/12. Hence, it is the correct answer.

The sample space for the given event is 2 * 2 * 2 = 8. The possible events when tossing a coin three times are, {HTT, TTT, THT, THH, TTH, HHH, HHT, HTH}. Here, there are three events in which even number of heads are there. Hence, the probability will be 3/8.

Here, the sample space is 21 and even numbers between 0 and 20 are {2, 4, 6, 8, 10, 12, 14, 16, 18}, i.e. 9 is the event value. Hence, the probability is 9/21.

Total marbles in the bowl = 6 + 5 = 11
The sample space is 11. Now, the probability of getting yellow marbles is 5/11. Hence, the probability of not getting yellow marbles is 1 - 5/11 = 6/11.

Probability of drawing a Queen is 4/52 = 1/13.
Since the card is replaced, the pack will again have 52 cards. So, the probability of drawing a King is 4/52 = 1/13.
Both events are independent, hence the probability of drawing both cards in succession is 1/13 * 1/13 = 1/169.

Here, the sample space is 52. The probability that the first card is queen = 4/52 and the probability that the second card is queen = 3/51. Hence, the probability that cards drawn are of Queen = 4/52 * 3/51 = 1/221.