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20 sandwiches are in a box. 11 of them are pb & j. what is the probabil…

Question

20 sandwiches are in a box. 11 of them are pb & j. what is the probability of drawing 2 pb & js if the first one is placed back in the box before the second draw? round your answer to the nearest hundredth.

Explanation:

Step1: Find probability of one PB&J

Probability = $\frac{\text{Number of PB&J}}{\text{Total sandwiches}} = \frac{11}{20} = 0.55$

Step2: Calculate probability of two draws

Since draws are with replacement, multiply probabilities: $0.55 \times 0.55 = 0.3025$

Step3: Round to nearest hundredth

$0.3025 \approx 0.30$

Answer:

0.30