Question

A die is cast until 6 appear. What is the probability that it must be cast more than five times?

Answer 1

1. Question⇒   A die is cast until 6 appear. What is the probability that it must be cast more than five times?

Step-by-step explanation:

• According to the question given, we have to throw a dice until 6 appears.
• And we have to find the probability that the dice is thrown more than 5 times.
• i.e. 6 must not appear in the Ist, 2nd, 3rd, 4th, and 5th throw.
• It would be easy to solve this question by negation. i.e. finding the probability that 6 appears either in 1st , 2nd, 3rd, 4th, or 5th throw. And then Subtracting this probability by 1, we will get our probability.
• Case 1: Probability of getting 6 in 1st throw:-
• P(1st) = 1/6
• Case 2: Probability of getting 6 in 2nd throw:-
• P(2nd) = 5/6 * 1/6
• Case 3: Probability of getting 6 in 3rd throw:-
• P(3rd) = 5/6 * 5/6 * 1/6
• Case 4: Probability of getting 6 in 4th throw:-
• P(4th) = 5/6 * 5/6 * 5/6 * 1/6
• Case 5: Probability of getting 6 in 5th throw:-
• P(5th) = 5/6 * 5/6 * 5/6 * 5/6 * 1/6.
• Now P(More than 5 times) = 1 - {P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th)}
• P(More than 5 times) = 1 - { 1/6 + 5/6*1/6 + (5/6)^2 *1/6 + (5/6)^3 *1/6 + (5/6)^4 *1/6}
• P(More than 5 times) = 1 - 1/6{1 + 5/6 + (5/6)^2 + (5/6)^3 + (5/6)^4}
• P(More than 5 times) = 1 - 1/6{(1 - (5/6)^5)/(1 - 5/6)} [Sum of n terms of a GP series.]
• P(More than 5 times) = 1 - {1 - (5/6)^5}
• P(More than 5 times) = 1 - 1 + (5/6)^5
• P(More than 5 times) = (5/6)^5
• Hence the required probability is (5/6)^5.

Answer 2

Answer:

The probability of appearing 6 more than 5 times is $(\frac{5}{6} )^{5}$

Step-by-step explanation:

Explanation :

Given , a die is cast until 6 appear .

6 must not appear in the 1st, 2nd, 3rd, 4th, and 5th throw.

We need to  find the probability that 6 appears either in 1st , 2nd, 3rd, 4th, or 5th throw.

Step 1:

Probability of getting 6 in 1st throw

⇒P(1st) = 1/6

Probability of getting 6 in 2nd throw

⇒P(2nd) = $\frac{5}{6} .\frac{1}{6}$

Probability of getting 6 in 3rd throw

⇒P(3rd) =$\frac{5}{6}. \frac{5}{6}. \frac{1}{5}$ = $(\frac{5}{6})^{2} (\frac{1}{6} )$

Probability of getting 6 in 4th throw

⇒P(4th) =  $\frac{5}{6} .\frac{5}{6} .\frac{5}{6} .\frac{1}{6}$ = $(\frac{5}{6} )^{3}$$. \frac{1}{6}$

Probability of getting 6 in 5th throw

⇒P(5th) = $\frac{5}{6}. \frac{5}{6} .\frac{5}{6} .\frac{5}{6} .\frac{1}{5}$ = $(\frac{5}{6} )^{4} .(\frac{1}{6} )$

Probability 6 appearing before the 6 throw =     {P(1st) + P(2nd) + P(3rd) + P(4th) + P(5th)

=  $[ \frac{1}{6} (1+(\frac{5}{6} )^{} +(\frac{5}{6} )^{2} +(\frac{5}{6} )^{3}+(\frac{5}{6} )^{4} )]$

⇒$\frac{1}{6} [\frac{1-(\frac{5}{6} )^{5} }{1-\frac{5}{6} } ]$  = $\frac{1}{6}[\frac{1- (\frac{5}{6} )^{5} }{\frac{1}{6} } ] = 1- (\frac{5}{6} )^{5}$  (Sum of n term of a GP series )

Now , P(More than 5 times) = 1 - (Probability 6 appearing before the 6 throw)

P(More than 5 times) =   $[1 - (1-(\frac{5}{6} )^{5} )]$

P(More than 5 times) = $[(\frac{5}{6} )^{5} ]$

Final answer :

Hence , the probability of appearing 6 more than 5 times is $(\frac{5}{6} )^{5}$.

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