a) A number card is drawn randomly from a group of number card numbered from 1 to 30. Find the probability of getting a card which is exactly divisible by 5 or 7. b) Thre are 20 toys of same shape…

a) Probability of a card divisible by 5 or 7

Step 1: Total number of possible outcomes The number cards are numbered from 1 to 30. Therefore, the total number of possible outcomes is: $$n(S)=30$$

Step 2: Find numbers divisible by 5 Numbers divisible by 5 between 1 and 30 are: $$5,10,15,20,25,30$$ Total numbers divisible by 5: $$n(A)=6$$

Step 3: Find numbers divisible by 7 Numbers divisible by 7 between 1 and 30 are: $$7,14,21,28$$ Total numbers divisible by 7: $$n(B)=4$$

Step 4: Find numbers divisible by both 5 and 7 (i.e., divisible by 35) Since 35 > 30, there are no numbers divisible by both 5 and 7 in this range: $$n(A\cap B)=0$$

Step 5: Calculate probability of divisible by 5 or 7 Using the formula for the probability of $A\cup B$: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ $$P(A\cup B)=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)}$$ $$P(A\cup B)=\frac{6}{30}+\frac{4}{30}-0=\frac{10}{30}=\frac{1}{3}$$

Final Answer: The probability of drawing a card divisible by 5 or 7 is $\frac{1}{3}$.

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b) Probabilities of all possible outcomes (Tree Diagram)

Step 1: Total number of toys

• Total toys = 20
• Red toys = 12
• Blue toys = 8

Step 2: Probabilities for first draw

• Probability of drawing a red toy first: $$P(R_1)=\frac{12}{20}=\frac{3}{5}$$
• Probability of drawing a blue toy first: $$P(B_1)=\frac{8}{20}=\frac{2}{5}$$

Step 3: Probabilities for second draw (without replacement)

• If first toy is red:

  • Remaining red toys = 11
  • Remaining blue toys = 8
  • Probability of drawing red second: $$P(R_2|R_1)=\frac{11}{19}$$
  • Probability of drawing blue second: $$P(B_2|R_1)=\frac{8}{19}$$

• If first toy is blue:

  • Remaining red toys = 12
  • Remaining blue toys = 7
  • Probability of drawing red second: $$P(R_2|B_1)=\frac{12}{19}$$
  • Probability of drawing blue second: $$P(B_2|B_1)=\frac{7}{19}$$

Step 4: Combined probabilities for all outcomes

• Probability of drawing two red toys: $$P(R_1\cap R_2)=P(R_1)\times P(R_2|R_1)=\frac{3}{5}\times \frac{11}{19}=\frac{33}{95}$$
• Probability of drawing one red and one blue toy (in any order): $$P(R_1\cap B_2)=\frac{3}{5}\times \frac{8}{19}=\frac{24}{95}$$ $$P(B_1\cap R_2)=\frac{2}{5}\times \frac{12}{19}=\frac{24}{95}$$
• Probability of drawing two blue toys: $$P(B_1\cap B_2)=\frac{2}{5}\times \frac{7}{19}=\frac{14}{95}$$

Tree Diagram:

Final Answer: The tree diagram above shows the probabilities of all possible outcomes when two toys are drawn in succession without replacement. The probabilities are:

• Two red toys: $\frac{33}{95}$
• One red and one blue toy (in any order): $\frac{24}{95}$ each
• Two blue toys: $\frac{14}{95}$