Question

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3, or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3, or 4 with the die?

Answer 1

Hints: To solve the question, at first we have to find out the probability of respective events that means the probability of getting 5 or 6 and getting 1,2,3 or 4 when the dice is thrown once. Then we must find out the probability of getting exactly one head when a coin is tossed thrice and the probability of getting exactly one head when a coin is tossed once. Finally, we apply Baye's theorem to find out the probability of getting 1, 2, 3, or 4 with the die when the coin is already tossed once and obtained exactly one head.

Complete step-by-step solution:
We know that the probability of an event A is given by the formula,
$p(A) = \dfrac{{n(A)}}{{n(S)}}$ …………………………………………….. (1)
Where n(A) is the number of outcomes obtained and n(S) is the total number of possible outcomes.
Let ${E_1}$be the event of getting 5 or 6 when a dice is thrown once. We see that when a dice is thrown once the possible outcomes are 1, 2,3,4,5 or 6 that means the total number of possible outcomes is 6. Here for ${E_1}$ the number of outcomes obtained are 2. Hence probability of event ${E_1}$ is given by
$p({E_1}) = \dfrac{2}{6} = \dfrac{1}{3}$ …………………………………………….. (2)
Let ${E_2}$ be the event of getting 1, 2, 3 or 4 when a dice is thrown thrice. Similarly for event ${E_2}$ the total number of outcomes obtained is 4 and total number of possible outcomes is 6. Hence probability of event${E_2}$is given by
$p({E_2}) = \dfrac{4}{6} = \dfrac{2}{3}$ …………………………………………….. (3)
Let A be the event of getting exactly one head. When a coin is tossed once then the total number of possible outcomes is 2 (H, T). When it is tossed thrice then the total number of possible outcomes is 8(HHH, HHT, HTH, THH, HTT, THT, TTH, TTT).
Let $p\left( {A\left| {{E_1}} \right.} \right)$ be the conditional probability of getting exactly one head when a coin is tossed thrice where event ${E_1}$ has already occurred. Here the number of outcomes obtained that contains is 3 (HTT, THT, TTH) out of 8 possible outcomes. Hence we get,
$p\left( {A\left| {{E_1}} \right.} \right) = \dfrac{3}{8}$ …………………………………………….. (4)
Again let $p\left( {A\left| {{E_2}} \right.} \right)$ be the conditional probability of getting a head when the coin is tossed once where the event ${E_2}$ is already occurred. Thus
$p\left( {A\left| {{E_2}} \right.} \right) = \dfrac{1}{2}$ …………………………………(5)
We know that the Baye's formula is given by,
$p\left( {A\left| B \right.} \right) = \dfrac{{p\left( {B\left| A \right.} \right)p(A)}}{{p\left( {B\left| A \right.} \right)p(A) + p\left( {B\left| {{A^c}} \right.} \right)p({A^c})}}$ …………………. (6)
Where
$p\left( {A\left| B \right.} \right)$ is the conditional probability that the event A will occur where B has already occurred. $p\left( {B\left| A \right.} \right)$ is the conditional probability that the event B will occur where B has already occurred.
${A^c}$ is the complementary event of A.
But here ${E_1}$ and ${E_2}$ are complementary to each other as out of 6 outcomes ${E_1}$ consists of 2 (5 or 6) and ${E_2}$ consists of 4 (1, 2, 3 or 4).
Hence by applying Baye's formula, and substituting the values from equations (2), (3), (4), (5) we get the probability of happening of event ${E_2}$ when the event A has already occurred is given by

p\left( {{E_2}\left| A \right.} \right) = \dfrac{{p\left( {A\left| {{E_2}} \right.} \right)p({E_2})}}{{p\left( {A\left| {{E_2}} \right.} \right)p({E_2}) + p\left( {A\left| {{E_1}} \right.} \right)p({E_1})}} \\\ = \dfrac{{\dfrac{2}{3} \times \dfrac{1}{2}}}{{\dfrac{1}{3} \times \dfrac{3}{8} + \dfrac{2}{3} \times \dfrac{1}{2}}} \\\ = \dfrac{1}{3} \times \dfrac{{23}}{{24}} \\\ = \dfrac{8}{{11}} \\\ \end{gathered} $$ **Thus we got that the probability that the girl threw 1, 2, 3 or 4 with the die is $$\dfrac{8}{{11}}$$.** **Note:** For determining the conditional probability Baye’s formula is useful. It states that probability of an event A is based on the sum of the conditional probabilities of event A given that event B has or has not occurred and the most important is that the events A and B must be independent, mathematically $$p(B) = p\left( {B\left| A \right.} \right)p(A) + p\left( {B\left| {{A^c}} \right.} \right)p({A^c})$$.