Probability Questions and Answers

Answer: A

The probability is \(\frac{48}{125}\).

Each match is won with probability 0.8 = \(\frac{4}{5}\) and not won (drawn or lost) with probability 0.2 = \(\frac{1}{5}\). Exactly two wins in three matches is a binomial count:

\[ \begin{aligned} P &= \binom{3}{2}\left(\tfrac{4}{5}\right)^{2}\left(\tfrac{1}{5}\right) \\ &= 3\cdot\tfrac{16}{25}\cdot\tfrac{1}{5} = \tfrac{48}{125} \end{aligned} \]

\(\frac{16}{125}\) forgets the factor 3 for the three possible positions of the non-win, and \(\frac{16}{25}\) is just the chance of two particular wins. \(\frac{25}{48}\) is not a probability of this form.

Answer: B

The most appropriate range is \(0.4 \le p \le 0.6\).

For any two events, \(P(S \cap T) \ge P(S) + P(T) - 1 = 0.8 + 0.6 - 1 = 0.4\), because the union cannot have probability above 1. And \(P(S \cap T) \le \min(0.8, 0.6) = 0.6\), since the intersection cannot exceed either event.

If the offers are independent, \(p = 0.48\), inside this range. Values below 0.4 are impossible, which rules out the first and third options, and values above 0.6 are impossible, ruling out the last.

Answer: D

There are six outcomes.

With two dice, the ordered pairs that sum to 7 are:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

That is six outcomes out of 36, which makes 7 the most likely total. Three is the number of unordered pairs, which forgets that (1, 6) and (6, 1) are different outcomes on two dice.

Answer: D

The probability is \(\frac{2}{7}\).

A leap year has 366 days, which is 52 full weeks plus 2 days. The 52 weeks give 52 Sundays for certain; a 53rd Sunday occurs only if one of the 2 extra days is a Sunday.

The two extra days form one of 7 equally likely pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Two of these contain a Sunday, so

\[ P = \frac{2}{7} \]

\(\frac{1}{7}\) would be the answer for an ordinary year of 365 days (one extra day). \(\frac{5}{7}\) is the chance of not getting 53 Sundays, and \(\frac{3}{4}\) and \(\frac{4}{7}\) match no count of the pairs.

Answer: A

The probability is \(\dfrac{75}{216}\).

Exactly one 6 in three rolls means one roll shows 6 (probability 1/6) and the other two do not (probability 5/6 each). The single 6 can come on any of the 3 rolls:

\[ \begin{aligned} P &= \binom{3}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2} \\ &= 3 \times \frac{25}{216} = \frac{75}{216} \end{aligned} \]

25/216 is the probability for one particular order, such as 6 on the first roll only, and forgets to multiply by the three positions. 1/6 is the chance of a 6 on a single roll, and 1/18 matches no case of this experiment.

Answer: D

The probability of a face card is \(\frac{3}{13}\).

Face cards are the jack, queen and king of each of the four suits, 3 × 4 = 12 cards in all. So:

\[ P = \frac{12}{52} = \frac{3}{13} \]

1/13 is the probability of one particular rank (say, a king), 2/13 would count only two ranks, and 1 would mean every card is a face card. Counting the aces as face cards, a common error, would give 16/52 = 4/13, which is not offered.

Answer: A

The probability is \(\frac{10}{19}\).

Sanjay gets no reply in two ways: his letter is lost, or it arrives and Kesari's reply is lost. Each letter is lost with probability 1/10.

\[ \begin{aligned} P(\text{no reply}) &= \tfrac{1}{10} + \tfrac{9}{10} \times \tfrac{1}{10} \\ &= \tfrac{19}{100} \\ P(\text{lost} \mid \text{no reply}) &= \tfrac{10/100}{19/100} \\ &= \tfrac{10}{19} \end{aligned} \]

\(\frac{1}{10}\) is the chance of loss before Sanjay knows there was no reply; the missing reply makes loss more likely. \(\frac{9}{100}\) is the chance that the letter arrived but the reply was lost, and \(\frac{9}{10}\) is the chance that a letter arrives.

Answer: D

A deterministic experiment gives the same result every time it is repeated under identical conditions - heating water to 100°C at sea level, for example.

A random experiment is the opposite: the conditions may be identical, yet the outcome varies from trial to trial. Probability theory is concerned with the second kind.

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 \( \tfrac{2}{5} \).

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: \( \tfrac{17}{30} \).