Question

A coach is training 3 players. He observes that the player A can hit a target 4 times in 5 shots, player B can hit 3 times in 4 shots and the player C can hit 2 times in 3 shots

From this situation answer the following:

1. Let the target is hit by A, B: the target is hit by B and, C: the target is hit by A and C. Then, the probability that A, B and, C all will hit, is

a. 4/5

b. 3/5

c. 2/5

d. 1/5

2. Referring to (i), what is the probability that B, C will hit and A will lose?

a. 1/10

b. 3/10

c. 7/10

d. 4/10

3. With reference to the events mentioned in (i), what is the probability that ‘any two of A, B and C will hit?

1. 1/30

2. 11/30

3. 17/30

4. 13/30

4. What is the probability that ‘none of them will hit the target’?

a. 1/30

b. 1/60

c. 1/15

d. 2/15

5. What is the probability that at least one of A, B or C will hit the target?

a. 59/60

b. 2/5

c. 3/5

d. 1/60

Answer 1

1. (c) 2/5

2. (a) 1/10

3. (d) 13/30

4. (b) 1/60

5. (a) 59/60

Answer 2

1. (c) \(\frac 25\)

P(all hit) = Probability A will hit x Probability B will hit x Probability C will hit

\(=\frac 45 \times \frac 34 \times \frac 23\)

\(=\frac 25\)

2. (a) \( \frac 1{10}\)

P(B and C will hit and A will lose) = Probability A will not hit x Probability B will hit x Probability C will hit

\(= (1 - \frac 45) \times \frac 34 \times \frac 23\)

\(= \frac 15 \times \frac 34 \times \frac 23\)

\(= \frac 1{10}\)

3. (d) \(\frac{13}{30}\)

P(any two will hit) = P(A will not hit) x P(B will hit) x P(C will hit)

+ P(A will hit) x P(B will not hit) x P(C will hit)

+ P(A will hit) x P(B will hit) x P(C will not hit)

4. (b) \(\frac 1{60}\)

P(none will hit) = Probability A will not hit x Probability B will not hit x Probability C will not hit

\(= (1 - \frac 45) \times (1 - \frac 34) \times (1 - \frac 23)\)

\(= \frac 15 \times \frac 14 \times \frac 13\)

\(= \frac 1{60}\)

5. (a) \( \frac {59}{60}\)

P(at least one will hit) = P(exactly one will hit) + P(exactly two will hit) + P(all will hit)

P(exactly one will hit)

P(exactly one will hit) = P(A will hit) P(B will not hit) x P(C will not hit)

+ P(A will not hit) x P (B will hit) x P (C will not hit)

+ P(A will not hit)x P (B will not hit) x P (C will hit)

P(exactly two will hit)

This is calculated in Question 3

P(exactly two will hit) = \(\frac{13}{30}\) 

P(all will hit)

This is calculated in Question 1

P(all will hit) = \(\frac 25\)

Now,

P(at least one will hit) = P(exactly one will hit) + P(exactly two will hit) + P(all will hit)

\(= \frac 9{60}+\frac{13}{30}+\frac 25\)

\(= \frac {59}{60}\)