Question:
Two dice are tossed 6 times. Then the probability that 7 will show an exactly four of the tosses is:
Two dice are tossed 6 times. Then the probability that 7 will show an exactly four of the tosses is:
Updated On: Jul 7, 2022
- $\frac{225}{18442}$
- $\frac{116}{20003}$
- $\frac{125}{15552}$
- $\frac{117}{17442}$
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The Correct Option is C
Solution and Explanation
This is the question of binomial distribution because dice tossed 6 times
we know, $P(x) = {^nC_r} p^r \, q^{n-r}$ where
p = prob of success
q = prob of failure
n = no. of toss
Now, Here n = 6,
p = prob of getting 7 = $\frac{1}{6}$
and q = 1 - p = 1 - $\frac{1}{6} = \frac{5}{6}$ , r = 4
$\therefore$ Required prob = ${^6C_4} \left( \frac{1}{6}\right)^4 \left( \frac{5}{6} \right)^2 = \frac{125}{15552}$
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Notes on Multiplication Theorem on Probability
Concepts Used:
Multiplication Theorem on Probability
In accordance with the multiplication rule of probability, the probability of happening of both the events A and B is equal to the product of the probability of B occurring and the conditional probability that event A happens given that event B occurs.
Let's assume, If A and B are dependent events, then the probability of both events occurring at the same time is given by:
\(P(A\cap B) = P(B).P(A|B)\)
Let's assume, If A and B are two independent events in an experiment, then the probability of both events occurring at the same time is given by:
\(P(A \cap B) = P(A).P(B)\)
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