Question

100 POINTS + BRAINLIEST !!!!!!!!
In a certain mathematics examination, the probability of being award a grade A is believed to be 0.3
For a group of 10 candidates, calculate:
i) the probability that at least half achieve grade A
The probability of being awarded a grade B is also 0.3
ii) Find the probability that all ten candidates achieve either grade A or grade B

Answer 1

Answer:  (i) 15%     (ii) 0.6%

Step-by-step explanation:

Given: P(A) = 0.3      ⇒ ~P(A) = 0.7

           P(B) = 0.3      ⇒ ~P(B) = 0.7

    P(A ∪ B) = 0.6      ⇒ ~P(A ∪ B) = 0.4

If you don't have a calculator function for the Binomial Formula, the equation is: $P(X)=\dfrac{n!}{X!(n-X)!}\cdot p^X\cdot (1-p)^{n-X}$     where

• n is the number of trials
• p is the probability of success
• X is the number of successes

(i) Probability of at least 5 --> P(X ≥ 5) means P(5) + P(6) + P(7) + P(8) + P(9) + P(10)

P(5): n = 10, p = 0.3, X = 5       → P(5) = 0.1029

P(6): n = 10, p = 0.3, X = 6       → P(6) = 0.0368

P(7): n = 10, p = 0.3, X = 7        → P(7) = 0.0090

P(8): n = 10, p = 0.3, X = 8        → P(8) = 0.0014

P(9): n = 10, p = 0.3, X = 9        → P(9) = 0.0001

P(10): n = 10, p = 0.3, X = 10     → P(10) = 0.0000  

                                                TOTAL = 0.1502    →   15.02%

(ii) Probability of A or B = 10 --> P(X = 10)

P(10): n = 10, p = 0.6, X = 10       → P(10) = 0.0060    →   0.6%