Answer:
a.) 0.4565
b.) 0.8656
c.) 0.4615
Step-by-step explanation:
We solve this using the probability distribution formula of combination.
nCr * p^r * q^n-r
Where
n = number of trials
r = successful trials
probability of success = p = 77% =0.77
Probability of failure= q = 1-0.77 = 0.23
a.) When exactly 3 opponents are defeated, When n = 3 and r = 3, probability becomes:
= 3C3 * 0.77³ * 0.23^0
= 1 * 0.456533 * 1
= 0.456533 = 0.4565 (4.d.p)
b.) When at least 2 opponents are defeated, that is when r = 2 and when r = 3,
When r = 2, probability becomes:
= 3C2 * 0.77² * 0.23¹
= 3 * 0.5929 * 0.23
= 0.409101
When 3 opponents are defeated, we calculated it earlier to be 0.456533
Hence, probability that at least 2 opponents are defeated
= 0.409101 + 0.456533
= 0.865634 = 0.8656(2.d.p)
c.) If 2 games are played, probability he defeat all 3 at least once in the game will be the sum (probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game) + (probability of defeating all three opponents in both games)
Probability of defeating all three opponents in the first game = 0.456533
Probability of not defeating all three opponents in the second game = 1 - 0.456533 = 0.543467
Hence ,
probability of defeating all 3 opponents in the first game and not defeating all 3 in the second game = 0.465633 * 0.543467 = 0.253056
probability of defeating all three opponents in both games
= 0.456533 * 0.456533
=0.208422
Probability he defeats all three opponents at least once in 2games
= 0.253056 + 0.208422
=0.461478 = 0.4615(4.d.p)
