| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Find unknown probability parameter |
| Difficulty | Standard +0.3 This is a standard tree diagram problem requiring setting up equations from given probabilities and solving for an unknown parameter, then applying conditional probability. While it involves multiple steps, the techniques are routine for S1 level—no novel insight required, just careful algebraic manipulation of probability rules. |
| Spec | 2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1-x) \times 0.9 + x \times 0.24 = 0.801\) | M1 | Equation with sum of two 2-factor probabilities \(= 0.801\) |
| A1 | Correct equation | |
| \(x = 0.15\) | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\geqslant 100 \text{ times given } \leqslant 3 \text{ views})\) | B1 | \(0.85 \times 0.1\) seen on its own as numerator or denominator of fraction |
| \(\frac{P(\geqslant 100 \text{ times} \cap \geqslant 3 \text{ views})}{P(\geqslant 3 \text{ views})}\) | M1 | Attempt at \(P(\geqslant 3 \text{ views})\) either \((0.85 \times p_1 + 0.15 \times p_2)\) or \(1 - 0.801\) seen anywhere |
| \(\frac{0.85 \times 0.1}{0.85 \times 0.1 + 0.15 \times 0.76}\) or \(1 - 0.801\) | A1 | Correct unsimplified \(P(\geqslant 3 \text{ views})\) as numerator or denominator of fraction |
| \(= 0.427\) | A1 [4] | Correct answer |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1-x) \times 0.9 + x \times 0.24 = 0.801$ | M1 | Equation with sum of two 2-factor probabilities $= 0.801$ |
| | A1 | Correct equation |
| $x = 0.15$ | A1 [3] | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\geqslant 100 \text{ times given } \leqslant 3 \text{ views})$ | B1 | $0.85 \times 0.1$ seen on its own as numerator or denominator of fraction |
| $\frac{P(\geqslant 100 \text{ times} \cap \geqslant 3 \text{ views})}{P(\geqslant 3 \text{ views})}$ | M1 | Attempt at $P(\geqslant 3 \text{ views})$ either $(0.85 \times p_1 + 0.15 \times p_2)$ or $1 - 0.801$ seen anywhere |
| $\frac{0.85 \times 0.1}{0.85 \times 0.1 + 0.15 \times 0.76}$ or $1 - 0.801$ | A1 | Correct unsimplified $P(\geqslant 3 \text{ views})$ as numerator or denominator of fraction |
| $= 0.427$ | A1 [4] | Correct answer |
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\includegraphics[max width=\textwidth, alt={}, center]{c0c7e038-805a-4237-a579-a6571b84f337-2_451_1530_1393_303}
A survey is undertaken to investigate how many photos people take on a one-week holiday and also how many times they view past photos. For a randomly chosen person, the probability of taking fewer than 100 photos is $x$. The probability that these people view past photos at least 3 times is 0.76 . For those who take at least 100 photos, the probability that they view past photos fewer than 3 times is 0.90 . This information is shown in the tree diagram. The probability that a randomly chosen person views past photos fewer than 3 times is 0.801 .\\
(i) Find $x$.\\
(ii) Given that a person views past photos at least 3 times, find the probability that this person takes at least 100 photos.
\hfill \mbox{\textit{CAIE S1 2015 Q4 [7]}}