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Home » 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 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 or 4 with the die?
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, \mathbf{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?
CBSE | Class 12 | Guides | Maths | NCERT | Probability
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, \mathbf{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?

Solution:

Let \mathrm{E} {1} be the event where the die outcome is 5 or6, \mathrm{E} {2} be the event where the die outcome is 1,2,3 or 4, and A be the event where the die outcome is exactly head.

Then P\left(E_{1}\right)=2 / 6=1 / 3

P\left(E_{2}\right)=4 / 6=2 / 3

We have eight alternatives if we toss a coin three times.

\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, THH, TTH, THT, HTT, TTT \}

\Rightarrow P\left(A \mid E_{1}\right)=P (obtaining exactly one head by tossing the coin three times if she get 5 or

6) =3 / 8

And \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P} (obtaining exactly one head by tossing the coin three times if she get 1 , 2,3 or 4) =1 / 2

Now, provided that she got exactly one head, the probability that the girl threw 1,2,3 or 4 with a die is \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right).

We have used Bayes’ theorem to arrive at our conclusion.

\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}

We may now retrieve the result by swapping the values.

=\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}}

=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}

=\frac{\frac{1}{3}}{\frac{3+8}{24}}=\frac{8}{11}

\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)=8 / 11

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