A die is thrown three times, the probability of getting a larger number than the previous number each time, is

Question

A die is thrown three times, the probability of getting a larger number than the previous number each time, is (A) (15/216) (B) (5/108) (C) (13/216) (D) None of these

Tardigrade Admin · Accepted Answer

Total number of exhaustive events =63=216. Now in the case of die 2>1 (i.e. second number is larger than first number). Let second number is i(i>1) then the first number can be selected by / - 1 ways and third number by 6-i ways so the number of ways in which three numbers can be choosen are (i-1) × 1 ×(6-i) So, the total number of favourable cases are =∑i=25(i-1)(6-i)=20 Hence, required probability is =(20/216)=(5/54)

Q. A die is thrown three times, the probability of getting a larger number than the previous number each time, is

$\frac{15}{216}$

$\frac{5}{108}$

$\frac{13}{216}$

None of these

Q. A die is thrown three times, the probability of getting a larger number than the previous number each time, is

21615​

1085​

21613​

None of these

$\frac{15}{216}$

$\frac{5}{108}$

$\frac{13}{216}$