Question

A and B draw from a purse containing 3 sovereigns and 4 shillings. Find their respective chances of first drawing a sovereign, the coins when drawn not being replaced.

Answer 1

For solving this problem, we find the probability of first drawing a sovereign by both A and B individually. We will consider all the cases of possible probability for the event. To compute the probability we will divide the favorable number of outcomes by the total number of outcomes. Also, the sum of favourable and not favourable outcomes is 1. Using this methodology, we can easily solve the problem.

Complete step-by-step answer:
According to the problem, we are given three sovereigns and four shillings and we are required to find the respective chances of A and B first drawing a sovereign.
Total number of sovereigns = 3
Total number of shillings = 4
Total number of coins = 7
Probability of A drawing a sovereign in first draw $=\dfrac{3}{7}\ldots (1)$
Now, to calculate the probability of A failing to draw a sovereign, we will use the equation,
Probability of favourable outcomes + Probability of unfavourable outcomes = 1
Therefore, subtracting the probability of A drawing a sovereign in first draw from 1, we get
Probability of A failing to drawing a sovereign in first draw $=\dfrac{4}{7}$
Since, A has drawn one coin so 6 coins are left, 3 sovereigns and 3 shillings. So, the total number of outcomes is now 6.
Now, Probability of B drawing a sovereign in first draw after A fails $=\dfrac{3}{6}$
Therefore, Probability that A fails to draw a sovereign and B draws a sovereign in his first attempt is the multiplications of probabilities of both events
$=\dfrac{4}{7}\cdot \dfrac{3}{6}=\dfrac{2}{7}\ldots (2)$
Now, if both A and B fail to draw a sovereign, then they pick one-one shillings in their respective draws. So, for A and B drawing shilling in the first attempt is choosing one shilling from seven and then another shilling from 6 coins. So, probability is
$=\dfrac{4}{7}\cdot \dfrac{3}{6}$
Now, the number of sovereign lefts is 3 out of 5 coins. So, in the next turn A will pick up sovereign. By using the above result, we get
Probability that A and B both fails to draw in first attempts respectively and then A it draws a sovereign $=\dfrac{4}{7}\cdot \dfrac{3}{6}\cdot \dfrac{3}{5}=\dfrac{6}{35}\ldots (3)$
Similarly, proceeding in the same manner until all the shillings are finished.
Probability that A fails twice and B fails once means three shillings are drawn before a sovereign. So, it can be expressed as $=\dfrac{4}{7}\cdot \dfrac{3}{6}\cdot \dfrac{2}{5}$
Now, B draws a sovereign from rest four coins amongst which 3 are sovereign and one is shilling $=\dfrac{4}{7}\cdot \dfrac{3}{6}\cdot \dfrac{2}{5}\cdot \dfrac{3}{4}=\dfrac{3}{35}\ldots (4)$
Probability that when both A and B fails twice means all four shillings are drawn before a sovereign. So, it can be expressed as $=\dfrac{4}{7}\cdot \dfrac{3}{6}\cdot \dfrac{2}{5}\cdot \dfrac{1}{4}$
Now, A draws a sovereign amongst left 3 coins $=\dfrac{4}{7}\cdot \dfrac{3}{6}\cdot \dfrac{2}{5}\cdot \dfrac{1}{4}\cdot \dfrac{3}{3}=\dfrac{1}{35}\ldots (5)$
Now, all the shillings are finished and hence the drawn coin will be sovereign on the next draw.
∴ chance of the winning of A is the sum of equation (1), (3) and (5).
$=\dfrac{3}{7}+\dfrac{6}{35}+\dfrac{1}{35}=\dfrac{22}{35}$
and the chance of the winning of B is the sum of equation (2) and (4).
$=\dfrac{2}{7}+\dfrac{3}{35}=\dfrac{13}{35}$
Hence, the chances of winning of A is $\dfrac{22}{35}$, and chances of winning of B is $\dfrac{13}{35}$.

Note: The obtained probabilities of chances of winning of A and B is correct because the sum of probabilities is 1. Students must be careful while making all the cases and must check at last whether the sum of the probabilities 1 to cross verify that his answer is correct. Since the denominator is the same and we have 22 and 13 in the numerator, we can add the numerators to save time. When adding we get 22 + 13 = 35. This is the same as the denominator, so it cancels and we get 1. So, it is correct.