Question

you play a game with 0.3 as the probability of winning.what is the probability that you win 4 or more games out of 10?(hint:round off to 3 decimal places)​

Answer 1

Step-by-step explanation:

Let A be an event of winning and B be an event of losing the game.

Given P(A)=0.3

Total probability =P(A)+P(B)

⇒ 1=0.3+P(B)

⇒ P(B)=1−0.3=0.7

∴ P(B)=0.7

hope it helps

Answer 2

Answer:

The probability of winning 4 or more games out of 10 is 0.060

Step-by-step explanation:

Let p be the probability of winning the game and q be the probability of losing the game.

The probability of winning is given to be 0.3, i.e., p = 0.3

We know that the sum of the probability of winning and the probability of losing is unity. Therefore,

p + q = 1

0.3 + q = 1

q = 1 - 0.3

q = 0.7

We have to find the probability of winning 4 or more games out of 10.

Therefore,

$P(X\geq 4) = \frac{4}{10} * \frac{5}{10} * \frac{6}{10} * \frac{7}{10} * \frac{8}{10} * \frac{9}{10} * \frac{10}{10}$

$P(X\geq 4) = \frac{604800}{10000000}$

$P(X\geq 4) = 0.0604800$

Rounding off to three decimal places, we get

P(X≥4) = 0.060

Therefore, the probability of winning 4 or more games is 0.060

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