Question

In an exam of three sections, the probability that Ronald will clear the three sections is 1/5, 2/3 and 3.4.
respectively. What is the probability that Ronald clears exactly one section?​

Answer 1

Answer:

7/20

Step-by-step explanation:

(1/5x1/3x1/4)+(4/5x2/3x1/4)+(3/4x1/3x4/5) = 1/60 + 9/60 + 12/60 = 21/60 = 7/20

Ronald passes one section in each of the below

(1/5x1/3x1/4) -   1/5 P, 1/3 F, 1/4 F

(4/5x2/3x1/4) - 4/5 F, 2/3 P, 1/4 F

(3/4x1/3x4/5) -  3/4 P, 1/3 F, 4/5 F

Answer 2

Given: In an exam of three sections, the probability that Ronald will clear the three sections is 1/5, 2/3 and 3/4 respectively.

To find: Probability that Ronald clears exactly one section

Explanation: In first case, let Ronald clear only the first section. So, he must fail the other sections.

Probability of clearing first section= 1/5

Probability of failing second section= 1-2/3 = 1/3

Probability of failing third section= 1-3/4 = 1/4

Combining all this, Probability of clearing only first section and failing other= 1/5 * 1/3 * 1/4

= 1/60

In second case, let Ronald clear only the second section. So, he must fail the other sections.

Probability of failing first section= 1-1/5= 4/5

Probability of clearing second section= 2/3

Probability of failing third section= 1-3/4 = 1/4

Combining all this, Probability of clearing only second section and failing other= 4/5 * 2/3 * 1/4

= 8/60

= 2/15

In third case, let Ronald clear only the third section. So, he must fail the other sections.

Probability of failing first section= 1-1/5= 4/5

Probability of failing second section= 1-2/3 = 1/3

Probability of clearing third section= 3/4 = 3/4

Combining all this, Probability of clearing only third section and failing other= 4/5 * 1/3 * 3/4

= 1/5

Adding all the probability to find the net probability to clear exactly one section,

P = 1/5 + 2/15 + 1/60

= 12+8+1/60

= 21/60

= 7/20

Therefore, the probability of clearing exactly one section is 7/20.