A girl throws a die. If she get 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 two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 OR 4 with the die?

Solution

Consider the following events:

E₁ = Getting 5 OR 6 in a single throw of the die

E₂ = Getting 1, 2, 3 OR 4 in a single throw of the die

A = Getting exactly 2 heads

We have to find, $P (E_{2} | A)$ .

$Since P (E_{2} | A) = \frac{P (A | E_{2}) P (E_{2})}{P (A | E_{1}) P (E_{1}) + P (A | E_{2}) P (E_{2})} Now, P (E_{1}) = \frac{2}{6} = \frac{1}{3} and P (E_{2}) = \frac{4}{6} = \frac{2}{3} Also, P (A | E_{1}) = Probability of getting exactly 2 heads when a coin is tossed 3 times = \frac{3}{8} And, P (A | E_{2}) = Probability of getting 2 heads when a coin is tossed 2 times = \frac{1}{2}$

$∴ P (E_{2} | A) = \frac{\frac{1}{2} \times \frac{2}{3}}{\frac{3}{8} \times \frac{1}{3} + \frac{1}{2} \times \frac{2}{3}} = \frac{\frac{1}{3}}{\frac{1}{3} (\frac{3}{8} + 1)} = \frac{\frac{1}{3}}{\frac{1}{3} (\frac{3}{8} + \frac{1 \times 8}{1 \times 8})} = \frac{\frac{1}{3}}{\frac{1}{3} (\frac{8 + 3}{8})} = \frac{\frac{1}{3}}{\frac{1}{3} (\frac{11}{8})} = \frac{1}{(\frac{11}{8})} = \frac{8}{11}$

For Daily Free Study Material Join wiredfaculty

Vector Algebra

A girl throws a die. If she get 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 two times and notes the number of heads obtained. If she obtained exactly two heads, what is the probability that she threw 1, 2, 3 OR 4 with the die?

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction