UGCNET Questions

When a fair coin is tossed twice, the sample space has four equally likely outcomes: HH, HT, TH and TT. Exactly one head occurs in the outcomes HT and TH. There are therefore two favorable cases out of four possible outcomes. The required probability is the ratio of favorable outcomes to total outcomes, which is 2/4 = 1/2.

Option A:
Option A, 1/4, would represent the probability of a single specific outcome like HH or TT, not the event of exactly one head. It counts only one case instead of both HT and TH.

Option B:
Option B correctly identifies two favorable outcomes and divides by the four equally likely possibilities. This simplifies to 1/2, representing a fifty percent chance of getting exactly one head in two tosses.

Option C:
Option C, 3/4, would mean that three of the four outcomes give exactly one head, but HH and TT clearly do not fit that description. It overestimates the probability.

Option D:
Option D, 2/3, does not emerge from the simple fraction 2/4 and would incorrectly suggest that only three outcomes are being considered. It cannot be justified from the enumerated sample space.

When three fair coins are tossed, there are 2³ = 8 equally likely outcomes. The easiest way to find the probability of at least one head is to subtract the probability of getting no heads (i.e., all tails) from 1. The probability of all tails is 1 outcome (TTT) out of 8, which is 1/8. Therefore, the probability of at least one head is 1 − 1/8 = 7/8.

Option A:
Option A, 1/8, gives the probability of the complementary event (no head) rather than at least one head. It represents the chance of all tails.

Option B:
Option B, 3/8, might come from counting only single-head outcomes and ignoring cases with two or three heads, so it underestimates the event.

Option C:
Option C, 5/8, still misses some favourable outcomes and does not match the simple complement calculation.

Option D:
Option D correctly uses the complement rule and recognizes that all outcomes except the single all-tail result satisfy “at least one head,” leading to 7/8.

A standard deck of playing cards has 52 cards divided into 4 suits: hearts, diamonds, clubs and spades, each with 13 cards. The favorable outcomes for drawing a heart are 13. Thus, the probability is the ratio of favorable outcomes to total outcomes, which is 13/52. This fraction simplifies to 1/4. So, the probability of drawing a heart is 1/4.

Option A:
Option A, 1/4, is exactly the simplified fraction 13/52. It acknowledges that there are four equally sized suits and that hearts constitute one of the four, giving a 25% chance on a single draw.

Option B:
Option B, 1/3, would require three suits or some other distribution of cards, which is not the case in a standard deck.

Option C:
Option C, 1/13, incorrectly treats the number of ranks as the total sample space, ignoring that each suit has all 13 ranks.

Option D:
Option D, 3/13, would imply that hearts comprise nearly one-quarter of the cards, but with a different count than the actual 13 per suit. It is not supported by the structure of the deck.