JEE Main 2020 — Probability Question with Solution

Number of ways 3 consecutive heads can appers

(1) HHHT_

(2) _THHH

(3) THHHT

$$\therefore$$ Probablity of getting 3 consecutive heads

= $$\frac{2}{32}$$ + $$\frac{2}{32}$$ + $$\frac{1}{32}$$ = $$\frac{5}{32}$$

Number of ways 4 consecutive heads can appers

(1) HHHHT

(2) THHHH

$$\therefore$$ Probablity of getting 4 consecutive heads

= $$\frac{1}{32}$$ + $$\frac{1}{32}$$ = $$\frac{2}{32}$$

Number of ways 5 consecutive heads can appers

(1) HHHHH

$$\therefore$$ Probablity of getting 5 consecutive heads

= $$\frac{1}{32}$$

Now Probablity of getting 0, 1, and 2 consecutive heads

= 1 - $$\left(\frac{5}{32}+\frac{2}{32}+\frac{1}{32}\right)$$ = $$\frac{24}{32}$$

Now, Expectation

= (-1) $$\times$$ $$\frac{24}{32}$$ + 3 $$\times$$ $$\frac{5}{32}$$ + 4 $$\times$$ $$\frac{2}{32}$$ + 5 $$\times$$ $$\frac{1}{32}$$

= $$\frac{1}{8}$$