Let Bn denotes the event that n fair dice are rolled once with P(Bn)=(1/2n), n ∈ N. e.g. P(B1)=(1/2), P(B2)=(1/22), P(B3)=(1/23), ldots ldots ldots and P(Bn)=(1/2n) Hence B 1, B 2, B 3, ldots ldots ldots B n are pairwise mutually exclusive and exhaustive events as n arrow ∞. The event A occurs with atleast one of the event B1, B2, ldots ldots ., Bn and denotes that the sum of the numbers appearing on the dice is S. Probability that greatest number on the dice is 4 if three dice are known to have been rolled, is

Question

Let Bn denotes the event that n fair dice are rolled once with P(Bn)=(1/2n), n ∈ N. e.g. P(B1)=(1/2), P(B2)=(1/22), P(B3)=(1/23), ldots ldots ldots and P(Bn)=(1/2n) Hence B 1, B 2, B 3, ldots ldots ldots B n are pairwise mutually exclusive and exhaustive events as n arrow ∞. The event A occurs with atleast one of the event B1, B2, ldots ldots ., Bn and denotes that the sum of the numbers appearing on the dice is S. Probability that greatest number on the dice is 4 if three dice are known to have been rolled, is (A)  (37/216) (B) (64/216) (C) (27/216) (D) (31/216)

Tardigrade Admin · Accepted Answer

n(S)=6 ⋅ 6 ⋅ 6=63=216 Now greatest number is 4 , so atleast one of the dice shows up 4 . ∴ n ( A )=43-33=37 Hence P(A)=(37/216)

$\frac{37}{216}$

$\frac{64}{216}$

$\frac{27}{216}$

$\frac{31}{216}$

$n (S) = 6 \cdot 6 \cdot 6 = 6^{3} = 216$
Now greatest number is 4 , so atleast one of the dice shows up 4 .
$∴ n (A) = 4^{3} - 3^{3} = 37$
Hence $P (A) = \frac{37}{216}$

21637​

21664​

21627​

21631​

n(S)=6⋅6⋅6=63=216 Now greatest number is 4 , so atleast one of the dice shows up 4 . ∴n(A)=43−33=37 Hence P(A)=21637​

$\frac{37}{216}$

$\frac{64}{216}$

$\frac{27}{216}$

$\frac{31}{216}$

$n (S) = 6 \cdot 6 \cdot 6 = 6^{3} = 216$
Now greatest number is 4 , so atleast one of the dice shows up 4 .
$∴ n (A) = 4^{3} - 3^{3} = 37$
Hence $P (A) = \frac{37}{216}$